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XVI OM - II - Task 6
Prove that there does not exist a polyhedron whose every planar section is a triangle. | Every vertex of a polyhedron is the common point of at least three of its edges. Let $ A $ be a vertex of the polyhedron, and segments $ AB $ and $ AC $ as well as $ AB $ and $ AD $ be edges of two faces. Let us choose points $ P $ and $ Q $ on segments $ AC $ and $ AD $, respectively, different from the endpoints of these segments, and draw through $ P $ and $ Q $ lines $ p $ and $ q $ parallel to line $ AB $. Lines $ p $ and $ q $ lie in the planes of two faces of the polyhedron, so there are segments $ p_1 $ and $ q_1 $ on them, belonging to these faces. The section of the polyhedron by the plane determined by lines $ p $ and $ q $ is not a triangle, since the boundary of this section contains two parallel segments $ p_1 $ and $ q_1 $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 755 |
XXXII - I - Problem 10
Determine all functions $ f $ mapping the set of all rational numbers $ \mathbb{Q} $ to itself that satisfy the following conditions:
a) $ f(1)=2 $,
b) $ f(xy) = f(x)f(y)-f(x+y)+1 $ for $ x, y \in \mathbb{Q} $. | If the function $ f $ satisfies conditions a) and b), then the function $ g: \mathbb{Q} \to \mathbb{Q} $ defined by the formula $ g(x) = f(x)- 1 $ satisfies the conditions
a') $ g(1) = 1 $
and $ g(xy)+1 = (g(x)+1)(g(y)+1)-g(x+y)-1+1 $, $ g(xy)+1 = g(x)g(y)+g(x)+g(y)+1-g(x+y) $, i.e.
b') $ g(xy)+g(x+y) = g(x)g(y)+g(x)+g(y) $.
Substituting $ y = 1 $ in the last equality, we get
thus, by a')
Hence
Substituting $ x = 0 $ and $ x = -1 $ in the last equality, we will determine, remembering a')
By obvious inductive reasoning, it follows from this and (*), that
for any integer $ n $, and
for an integer $ m $ and a rational $ x $.
Now let $ m $ be an integer, and $ n $ a natural number.
To determine $ g\left( \frac{m}{n} \right) $, substitute $ x = m $, $ y = \frac{1}{n} $ in equality b).
Since $ g(m)= m $ and $ g\left( m+\frac{1}{n} \right)=m+g \left( \frac{1}{n} \right) $, we have
If $ m = n $, then $ 1 = g(1) = n \cdot g \left( \frac{1}{n} \right) $, from which it follows that
Therefore
This means that if the function $ g $ satisfies conditions a') and b'), then $ g(x) = x $ for every rational $ x $. Therefore, if the function $ f $ satisfies conditions a) and b), then $ f(x) = x+ 1 $. On the other hand, the function $ f $ defined by this formula obviously satisfies conditions a) and b). The only solution to the problem is therefore the function $ f: \mathbb{Q} \to \mathbb{Q} $, defined by the formula $ f(x) = x+1 $. | f(x)=x+1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 757 |
LVIII OM - II - Problem 1
The polynomial $ P(x) $ has integer coefficients. Prove that if the polynomials $ P(x) $ and $ P(P(P(x))) $ have a common real root, then they also have a common integer root. | Let the real number $ a $ be a common root of the polynomials $ P(x) $ and $ P(P(P(x))) $. From the equalities $ P(a)=0 $ and $ P(P(P(a)))=0 $, we obtain $ P(P(0))=0 $. The integer $ m=P(0) $ thus satisfies the conditions
which means that $ m $ is the desired common integer root. | proof | Algebra | proof | Yes | Yes | olympiads | false | 759 |
XLII OM - III - Problem 5
Non-intersecting circles $ k_1 $ and $ k_2 $ lie one outside the other. The common tangents to these circles intersect the line determined by their centers at points $ A $ and $ B $. Let $ P $ be any point on circle $ k_1 $. Prove that there exists a diameter of circle $ k_2 $, one end of which lies on the line $ PA $, and the other end on the line $ PB $. | Let the centers of circles $k_1$ and $k_2$ be denoted by $O_1$ and $O_2$, and their radii by $r_1$ and $r_2$. Points $A$ and $B$ lie on the line $O_1O_2$, and it does not matter which letter denotes which point. For the sake of clarity, let us agree that point $B$ lies on the segment $O_1O_2$, and point $A$ lies outside this segment.
Consider one of the common tangents to circles $k_1$ and $k_2$ passing through point $A$, and one of the common tangents passing through point $B$. The points of tangency of the first line with circles $k_1$ and $k_2$ are denoted by $U_1$ and $U_2$, and those of the second line by $V_1$ and $V_2$ (Figure 9 shows the situation when $r_1 < r_2$; but the entire subsequent reasoning does not change when $r_1 > r_2$ - we recommend the Reader to draw the corresponding figure).
om42_3r_img_9.jpg
From the similarity of right triangles $AU_1O_1$ and $AU_2O_2$ and the similarity of triangles $BV_1O_1$ and $BV_2O_2$, the following proportions follow:
Let $j_A$ be the homothety with center $A$ and ratio $r_2/r_1$; let $j_B$ denote the homothety with center $B$ and ratio $-r_2/r_1$. From the proportions obtained just now, it follows that
We see, therefore, that the image of circle $k_1$ under each of these homotheties is circle $k_2$.
Consider the points
which lie on circle $k_2$. Point $R$ is the image of point $Q$ under the transformation
The inverse transformation $j_A^{-1}$ (inverse to $j_A$) is a homothety with ratio $r_1/r_2$. Thus, $f$ is an isometry, and an even isometry (as the composition of two homotheties).
Circle $k_2$ is transformed by $j_A^{-1}$ into circle $k_1$: circle $k_1$ is transformed by $j_B$ into circle $k_2$. Therefore, the image of circle $k_2$ under transformation $f$ is the same circle $k_2$. Hence, $f$, as an even isometry, is a rotation about point $O_2$ by some angle.
To determine this angle, let $C_1$ and $D_1$ be the points of intersection of line $O_1O_2$ with circle $k_1$, and $C_2$ and $D_2$ the points of intersection of this line with circle $k_2$. Let us choose the notation so that the considered points lie on line $O_1O_2$ in the order: $C_1$, $O_1$, $D_1$, $B$, $D_2$, $O_2$, $C_2$. Then $j_A(D_1) = C_2$, $j_B(D_1) = D_2$, and thus $f(C_2) = D_2$.
Segment $C_2D_2$ is a diameter of circle $k_2$; rotation $f$ maps one of its endpoints to the other. Therefore, $f$ is a rotation by $180^\circ$.
We previously noted that $f$ maps point $Q$ to $R$. This means that segment $QR$ is also a diameter of circle $k_2$. Point $Q$ lies on line $PA$, and point $R$ on line $PB$. Therefore, segment $QR$ is a diameter, whose existence we needed to prove. | proof | Geometry | proof | Yes | Yes | olympiads | false | 761 |
XII OM - II - Task 4
Find the last four digits of the number $ 5^{5555} $. | \spos{1} We will calculate a few consecutive powers of the number $ 5 $ starting from $ 5^4 $:
It turned out that $ 5^8 $ has the same last four digits as $ 5^4 $, and therefore the same applies to the numbers $ 5^9 $ and $ 5^5 $, etc., i.e., starting from $ 5^4 $, two powers of the number $ 5 $, whose exponents differ by a multiple of $ 4 $, have the same last four digits. The number $ 5^{5555} = 5^{4\cdot 1388+3} $ therefore has the same last $ 4 $ digits as the number $ 5^7 $, i.e., the digits $ 8125 $. | 8125 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 766 |
XLV OM - I - Problem 6
The function $ f: \mathbb{R} \to \mathbb{R} $ is continuous. Prove that if for every real number $ x $ there exists a natural number $ n $ such that $ (\underbrace{f \circ \ldots \circ f}_{n})(x) = 1 $, then $ f(1)=1 $. | We will denote by $ f^r $ the $ r $-th iterate of the function $ f $, that is, the $ r $-fold composition (superposition) $ f \circ f \circ \ldots \circ f $.
Suppose, contrary to the thesis, that $ f(1) \neq 1 $. Let $ m $ be the smallest positive integer for which the equality
holds (such a number exists by the condition of the problem). Since $ f(1) \neq 1 $, we have $ m \geq 2 $. Let $ x_0 = 1 $ and denote the value $ f(1) $ by $ x_1 $, the value $ f(x_1) $ by $ x_2 $, and so on, inductively, the value $ f(x_i) $ by $ x_{i+1} $ (for $ i = 1,2,\ldots,m - 1 $); then $ x_m = f^m(1) = 1 $. We obtain a sequence of real numbers $ x_0,x_1,\ldots,x_{m-1},x_m $ satisfying the conditions
Let $ x_k $ be the smallest, and $ x_l $ the largest among the numbers $ x_0,x_1, \ldots ,x_{m-1} $. Therefore, the inequalities
hold (indeed, strict inequalities hold here because the numbers $ x_0, x_1, \ldots, x_{m-1} $ are all different, which follows from the definition of $ m $ as the smallest number satisfying condition (1); but for the further reasoning, weak inequalities (2) are sufficient).
Consider the function $ g(x) = f(x) - x $; this is a continuous function. It takes the values
at the points $ x_k $ and $ x_l $
and
therefore, by the Darboux property, there exists a number $ c \in \langle x_k; x_l \rangle $ for which the equality $ g(c) = 0 $ holds, i.e., $ f(c) = c $. This implies that
Let us use the condition of the problem again, which states that for some $ n $ the value $ f^n(c) $ is equal to $ 1 $. We conclude that $ c = 1 $. The equality $ f(c) = c $ now asserts that $ f(1) = 1 $ - contrary to the initial assumption - and thus proves the incorrectness of this assumption. Therefore, ultimately, $ f(1)=1 $. | proof | Algebra | proof | Yes | Yes | olympiads | false | 767 |
XLVII OM - III - Problem 1
Determine all pairs $ (n,r) $, where $ n $ is a positive integer and $ r $ is a real number, for which the polynomial $ (x + 1)^n - r $ is divisible by the polynomial $ 2x^2 + 2x + 1 $. | Let $ Q_n(x) $ and $ R_n(x) $ denote the quotient and remainder, respectively, when the polynomial $ (x + 1)^n $ is divided by $ 2x^2 + 2x + 1 $. The pair $ (n,r) $ is one of the sought pairs if and only if $ R_n(x) $ is a constant polynomial, identically equal to $ r $.
For $ n = 1,2,3,4 $ we have:
Therefore,
For every integer $ n \geq 0 $, the following equality holds:
Hence, $ R_{n+4}(x) = -\frac{1}{4}R_n(x) $. From this and the formulas (1), we obtain by induction the following equalities (for $ k = 1,2,3,\ldots $):
As we can see, $ R_n(x) $ is a constant polynomial only for $ n = 4k $; its constant value is then $ ( -\frac{1}{4})^k $.
Conclusion: the sought pairs $ (n,r) $ have the form $ (4k,(-\frac{1}{4})^k) $, where $ k $ is a positive integer. | (4k,(-\frac{1}{4})^k) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 770 |
XVII OM - I - Problem 6
Prove the theorem: If the coefficients $ a, b, c, d $ of the cubic equation $ ax^3 + bx^2 + cx + d = 0 $ are integers, where the number $ ad $ is odd and the number $ bc $ is even, then the equation cannot have three rational roots. | Suppose the given equation has three rational roots $x_1, x_2, x_3$. Let us transform the equation by introducing a new variable $y = ax$, i.e., substituting $\frac{y}{a}$ for $x$ in the equation.
We obtain the equation
whose roots are $y_1 = ax_1$, $y_2 = ax_2$, $y_3 = ax_3$. These numbers, being the products of the rational numbers $x_1$, $x_2$, $x_3$ by the integer $a$, are rational. Since in equation (1) the coefficient of the highest power of the unknown, i.e., of $y^3$, is equal to $1$, it follows from a known theorem that the rational roots of this equation, i.e., $y_1$, $y_2$, $y_3$, are integers.
According to Vieta's formulas,
From the assumption that the product $ad$ of the integers $a$ and $d$ is an odd number, it follows that $a$ and $d$, and therefore $a^2d$, are odd numbers, from which, in view of equation (4), we conclude that $y_1$, $y_2$, $y_3$ are odd. In this case, from equations (2) and (3), it follows that the numbers $b$ and $ac$ are odd, and hence the product $abc$ is an odd number. We have thus obtained a contradiction with the assumption that the number $bc$ is even. The assumption that the given equation has three rational roots is therefore false. | proof | Algebra | proof | Yes | Yes | olympiads | false | 772 |
LV OM - I - Task 1
Given is a polygon with sides of rational length, in which all internal angles are equal to $ 90^{\circ} $ or $ 270^{\circ} $. From a fixed vertex, we emit a light ray into the interior of the polygon in the direction of the angle bisector of the internal angle at that vertex. The ray reflects according to the principle: the angle of incidence is equal to the angle of reflection. Prove that the ray will hit one of the vertices of the polygon. | Since every internal angle of the considered polygon $ \mathcal{W} $ is $ 90^{\circ} $ or $ 270^{\circ} $, all sides of this polygon (after an appropriate rotation) are aligned horizontally or vertically (Fig. 1). Let $ p_1/q_1, p_2/q_2, \ldots, p_n/q_n $ be the lengths of the consecutive sides of the polygon $ \mathcal{W} $, where each of the numbers $ p_1, p_2, \ldots, p_n $, $ q_1, q_2, \ldots, q_n $ is a positive integer.
Consider a square grid where each small square has a side length of $ 1/(q_1 q_2 \ldots q_n) $. Then the polygon $ \mathcal{W} $ can be placed on this grid such that each of its sides lies on a line defining the grid (Fig. 1). A light ray emitted from a vertex of the polygon $ \mathcal{W} $ according to the given rules moves along the diagonals of the grid squares and reflects off the sides of the polygon $ \mathcal{W} $ at grid points.
om55_1r_img_1.jpg
Suppose the light ray did not hit any vertex of the polygon $ \mathcal{W} $. It therefore reflected off the sides of the polygon an arbitrary number of times. This implies that the light ray hit a point $ \mathcal{P} $ on the boundary of the polygon $ \mathcal{W} $ at least three times. Thus, the light ray hit the point $ \mathcal{P} $ at least twice moving in the same direction. This means that the trajectory of the light ray is periodic, which is not possible, since the light ray started its journey from a vertex of the polygon $ \mathcal{W} $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 773 |
XLIX OM - I - Problem 7
Determine whether there exists a convex polyhedron with $ k $ edges and a plane not passing through any of its vertices and intersecting $ r $ edges, such that $ 3r > 2k $. | Let $ W $ be any convex polyhedron; let $ k $ be the number of its edges, and $ s $ -- the number of faces. The boundary of each face contains at least three edges, and each edge is a common side of exactly two faces. Therefore, $ 2k \geq 3s $.
If a plane not passing through any vertex intersects $ r $ edges, then the section of the polyhedron by this plane is a convex polygon with $ r $ vertices, and thus also $ r $ sides. Each of its sides is contained in some face of the polyhedron $ W $, and each in a different one. Therefore, $ s \geq r $.
Combining the obtained inequalities, we see that $ 2k \geq 3r $. Therefore, there does not exist a polyhedron for which the inequality $ 3r > 2k $ would be satisfied. | proof | Geometry | proof | Yes | Yes | olympiads | false | 774 |
IX OM - I - Problem 12
Given a circle with radius $ r $ and rays $ AB $ and $ AC $ tangent to this circle at points $ B $ and $ C $. Determine the tangent to the given circle such that the segment contained within the angle $ BAC $ is the shortest. | The desired straight line should of course be sought among the tangents to the given circle that intersect the segments $AB$ and $AC$, because for other tangents, the segments contained in the angle $BAC$ are larger than the segments of the tangents parallel to them. Let $MN$ (Fig. 13) be the segment of the tangent line to the given circle. Consider the triangle $MON$. The angle $MON$ of this triangle equals half of $\measuredangle BOC$, because $\measuredangle MOH = \measuredangle MOB$, and $\measuredangle HON = \measuredangle NOC$; since $\measuredangle BOC = 180^\circ - \measuredangle A$, therefore $\measuredangle MON = 90^\circ - \frac{1}{2} \measuredangle A$. The height $OH$ of the triangle $MON$ equals the radius $r$ of the given circle. Thus, in the triangle $MON$, the angle at vertex $O$ and the height from this vertex have a constant size, and the problem reduces to the following: which of the triangles with a given height and a given angle at the vertex has the smallest base? To answer this question, it is sufficient to rely on the construction of a triangle when we have given: the base $l$, the angle at the vertex $\alpha$, and the height $h$. This known construction is shown in Fig. 14. When $h$ and $\alpha$ remain unchanged, and $l$ decreases, then the radius $MS$ of the arc containing the angle $\alpha$ also decreases. The length $l$ is the smallest when this arc is tangent to the line $KL$. The triangle is then isosceles. Therefore, the sought straight line is the tangent to the circle for which $\triangle MON$ is isosceles, i.e., which is perpendicular to the axis of symmetry $AO$ of the given figure. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 775 |
XXXVIII OM - II - Zadanie 4
Wyznaczyć wszystkie pary liczb rzeczywistych $ a, b $, dla których wielomiany $ x^4 + 2ax^2 + 4bx + a^2 $ i $ x^3 + ax - b $ mają dwa różne wspólne pierwiastki rzeczywiste.
|
Oznaczmy rozważane wielomiany przez $ P $ i $ Q $:
Załóżmy, że wielomiany $ P $ i $ Q $ mają wspólne pierwiastki rzeczywiste $ x_1 $, $ x_2 $ ($ x_1\neq x_2 $). Liczby $ x_1 $ $ x_2 $ są wówczas także pierwiastkami wielomianu
Stąd $ a \neq 0 $ i $ \Delta = 9b^2--4a^3 > 0 $.
Zgodnie z wzorami Viete'a
Ponieważ $ Q(x_1) = Q(x_2) = 0 $, a $ x_1 \neq x_2 $, zachodzą równości
Zatem $ b = 0 $ i $ 0 < \Delta = -4a^3 $, czyli $ a < 0 $.
Także i odwrotnie: jeśli spełnione są warunki $ a < 0 = b $, to wielomiany $ P $ i $ Q $ przybierają postać
i liczby $ \sqrt{-a} $, $ -\sqrt{-a} $ są ich wspólnymi pierwiastkami.
Rozwiązaniem zadania są więc wszystkie pary postaci $ (a, 0) $, gdzie $ a < 0 $.
| (,0),where0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 779 |
XXXIV OM - III - Problem 5
In the plane, vectors $ \overrightarrow{a_1}, \overrightarrow{a_2}, \overrightarrow{a_3} $ of length 1 are given. Prove that one can choose numbers $ \varepsilon_1, \varepsilon_2, \varepsilon_3 $ equal to 1 or 2, such that the length of the vector $ \varepsilon_1\overrightarrow{a_1} + \varepsilon_2\overrightarrow{a_2} + \varepsilon_3\overrightarrow{a_3} $ is not less than 2. | Let's start from an arbitrary point $0$ with vectors $\overrightarrow{a_1}$, $\overrightarrow{a_2}$, $\overrightarrow{a_3}$. They determine three lines intersecting at point $0$. Among the angles determined by pairs of these lines, there is an angle of measure $\alpha$ not greater than $\frac{\pi}{3}$. Suppose this is the angle between the lines determined by vectors $\overrightarrow{a_1}$ and $\overrightarrow{a_2}$ (otherwise, we can change the numbering of the vectors). Therefore, one of the angles $\measuredangle (\overrightarrow{a_1}, \overrightarrow{a_2})$, $\measuredangle (\overrightarrow{a_1}, -\overrightarrow{a_2})$ has a measure $\alpha \in \left[0, \frac{\pi}{3} \right]$.
We choose $\varepsilon_2$ so that the angle $\measuredangle (\overrightarrow{a_1}, \varepsilon_2 \overrightarrow{a_2})$ has a measure $\alpha$. Let's compute the dot product of the vector $\overrightarrow{b} = \overrightarrow{a_1} + \varepsilon_2 \overrightarrow{a_2}$ with itself:
Therefore, $|\overrightarrow{b}| \geq \sqrt{3}$.
Next, notice that
Therefore, one of the numbers $(\overrightarrow{a_1} + \varepsilon_2 \overrightarrow{a_2} + \overrightarrow{a_3})^2$, $(\overrightarrow{a_1} + \varepsilon_2 \overrightarrow{a_2} - \overrightarrow{a_3})^2$ is not less than $4$. We choose $\varepsilon_3$ so that $(\overrightarrow{a_1} + \varepsilon_2 \overrightarrow{a_2} + \varepsilon_3 \overrightarrow{a_3})^2 \geq 4$. Then, $|\overrightarrow{a_1} + \varepsilon_2 \overrightarrow{a_2} + \varepsilon_3 \overrightarrow{a_3}| \geq 2$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 782 |
XXV OM - II - Problem 3
Prove that the orthogonal projections of vertex $ D $ of the tetrahedron $ ABCD $ onto the bisecting planes of the internal and external dihedral angles at edges $ \overline{AB} $, $ \overline{BC} $, and $ \overline{CA} $ lie on the same plane. | The bisecting plane is a plane of symmetry of the dihedral angle. Therefore, the image $D'$ of the vertex $D$ in the symmetry relative to any of the considered bisecting planes lies in the plane $ABC$. It follows that if $P$ is the orthogonal projection of the point $D$ onto the bisecting plane, then $P$ is the midpoint of the segment $\overline{DD'}$. Thus, the point $P$ is the image of the point $D$ in the homothety $\varphi$ with the center at point $D$ and the ratio $\displaystyle \frac{1}{2}$. Therefore, the projections of the point $P$ onto all the considered bisecting planes lie in the plane which is the image of the plane $ABC$ in the homothety $\varphi$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 783 |
XXXV OM - II - Problem 2
On the sides of triangle $ABC$, we construct similar isosceles triangles: triangle $APB$ outside triangle $ABC$ ($AP = PB$), triangle $CQA$ outside triangle $ABC$ ($CQ = QA$), and triangle $CRB$ inside triangle $ABC$ ($CR = RB$). Prove that $APRQ$ is a parallelogram or that points $A, P, R, Q$ lie on a straight line. | Consider a similarity with a fixed point $C$ that transforms $B$ into $R$ (this is the composition of a rotation around $C$ by the angle $BCR$ and a homothety with center $C$ and scale equal to the ratio of the base length to the arm length in each of the constructed isosceles triangles). This similarity transforms $A$ into $Q$. It follows that triangles $ABC$ and $QRC$ are similar.
om35_2r_img_8.jpg
Similarly, we observe that a similarity with a fixed point $B$ that transforms $C$ into $R$ transforms $A$ into $P$, so triangles $ABC$ and $PBR$ are similar. Therefore, triangles $QRC$ and $PBR$ are similar. Since $BR = CR$, these triangles are congruent, and thus $QC = PR$ and $QR = PB$. But $QC = AQ$ and $PB = AP$, so $AQ = PR$ and $AP = QR$. Moreover, the angle between vectors $\overrightarrow{BP}$ and $\overrightarrow{RQ}$ has a measure equal to $\measuredangle PBA + \measuredangle BCR = 2 \measuredangle PBA$, and the angle between vectors $\overrightarrow{AP}$ and $\overrightarrow{BP}$ has a measure equal to $180^\circ - 2 \measuredangle PBA$, which implies that vectors $\overrightarrow{RQ}$ and $\overrightarrow{AP}$ are parallel. Therefore, points $A$, $P$, $R$, $Q$ lie on the same line or are consecutive vertices of a parallelogram. | proof | Geometry | proof | Yes | Yes | olympiads | false | 784 |
XIII OM - I - Problem 3
Prove that the perpendiculars dropped from the centers of the excircles of a triangle to the corresponding sides of the triangle intersect at one point. | A simple solution to the problem can be inferred from the observation that on each side of the triangle, the point of tangency of the inscribed circle and the point of tangency of the corresponding excircle are symmetric with respect to the midpoint of that side. For example, if the inscribed circle in triangle $ABC$ touches side $BC$ at point $K$, and the excircle at point $L$, then, using the usual notation for the sides of the triangle, we have $CK = \frac{1}{2} (a + b - c)$, $BL = \frac{1}{2} (a + b - c)$, so $CK = BL$, i.e., points $K$ and $L$ are symmetric with respect to the midpoint $M$ of side $BC$ (Fig. 3).
Let $S$ be the center of the inscribed circle in triangle $ABC$, $O$ - the center of the circumscribed circle of this triangle, and $T$ the point symmetric to point $S$ with respect to point $O$. Since $SO = OT$, $KM = ML$, and lines $SK$ and $OM$ are parallel, line $TL$ is symmetric to line $SK$ with respect to the axis $OM$. Therefore, line $TL$ is perpendicular to side $BC$ at the point of tangency with the excircle, and thus passes through the center of this circle. The same reasoning applies to sides $AB$ and $AC$. Hence, it follows that through point $T$ pass the perpendiculars drawn from the centers of the excircles to the corresponding sides of the triangle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 788 |
XLV OM - III - Task 4
We have three unmarked vessels: an empty $ m $-liter, an empty $ n $-liter, and a full $(m+n)$-liter vessel of water. The numbers $ m $ and $ n $ are relatively prime natural numbers. Prove that for every number $ k \in \{1,2, \ldots , m+n-1\} $, it is possible to obtain exactly $ k $ liters of water in the third vessel by pouring water. | When pouring water from one container to another, we either completely empty the container we are pouring from or completely fill the container we are pouring into. (Since the containers have no markings, this is the only method that allows us to control the volume of water being poured.) Thus, after each pour, we have a non-negative integer number of liters of water in each container (not exceeding, of course, the capacity of the given container). Let's denote:
$ A $ - the $ m $-liter container,
$ B $ - the $ n $-liter container,
$ C $ - the $ (m+n) $-liter container.
Suppose that at some point, container $ C $ contains $ c $ liters of water. We will show that with at most two pours, we can achieve a state where container $ C $ contains $ c $ liters of water, where
Let $ a $ and $ b $ be the volume (in liters) of water in containers $ A $ and $ B $ at the considered moment. Thus, $ a $, $ b $, and $ c $ are integers,
Assume that $ c \leq m $. Then $ a + b \geq n $. We pour water from container $ A $ to container $ B $, completely filling it; that is, we pour $ n - b $ liters from $ A $ to $ B $ (this is possible because $ a \geq n - b $). Then we pour the entire contents of container $ B $, which is $ n $ liters, into $ C $, obtaining $ c + n $ liters in container $ C $.
Consider the second case: $ c > m $. Then $ a + b < n $. We pour the entire contents of container $ A $, which is $ a $ liters, into $ B $. We pour $ m $ liters from $ C $ into the empty container $ A $; in container $ C $, $ c - m $ liters of water remain.
Thus, we have obtained $ c $ liters in container $ C $, where the number $ c $ is given by formula (1). From this formula, it follows that
At the initial moment, container $ C $ contained $ m + n $ liters. By applying the described procedure, we obtain in container $ C $:
\begin{center}
after (at most) two pours - $ c_1 $ liters;\\
after (at most) four pours - $ c_2 $ liters;\\
etc.;\\
after (at most) $ 2i $ pours - $ c_i $ liters,
\end{center}
where
We will prove that every number $ k \in \{1,2,\ldots,m+n-1\} $ is equal to $ c_i $ for some $ i $; it is therefore a possible volume of water in container $ C $.
Let $ c_0 = 0 $. Each of the numbers
belongs to the set $ \{0,1,2,\ldots,m+n-1\} $; it is, according to formula (2), the remainder of the division of the product $ in $ by $ m + n $.
The numbers $ m $ and $ n $ are relatively prime by assumption. Therefore, the numbers $ n $ and $ m + n $ are also relatively prime. It follows that the remainders (3) are all different. Indeed: if for some numbers $ i $, $ j $ such that $ 0 \leq i < j \leq m + n - 1 $ the equality $ c_i = c_j $ held, it would mean that the difference $ jn - in $, i.e., the product $ (j - i)n $, is divisible by $ m + n $. The second factor of this product (i.e., $ n $) has no common prime factors with the number $ m + n $; this number must therefore be a divisor of the difference $ (j - i) $ - a contradiction, since $ 0 < j - i < m + n $.
Thus, indeed, the remainders (3) listed above are different elements of the set $ \{0,1,2,\ldots, m+n-1\} $. There are $ m + n $ of them. Therefore, every number $ k $ belonging to this set is equal to some number $ c_i $. This is precisely what we intended to prove.
Note: The described method allows achieving any given value $ k $ in no more than $ 2(m + n -1) $ pours. This estimate can be improved. If the desired number $ k $ is in the second half of the sequence (3), then using the "reverse" procedure, which with at most two pours will cause the volume of water in container $ C $ to change from value $ c $ to value
We leave it to the Reader as an exercise to refine the details and convince themselves that by choosing one of these procedures, we can achieve the desired value $ k $ in no more than $ 2[\frac{1}{2}(m + n)] $ pours. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 789 |
XLVI OM - II - Problem 3
Given are positive irrational numbers $ a $, $ b $, $ c $, $ d $, such that $ a+b = 1 $. Prove that $ c+d = 1 $ if and only if for every natural number $ n $ the equality $ [na] +[nb] = [nc] + [nd] $ holds.
Note: $ [x] $ is the greatest integer not greater than $ x $. | Let's take any natural number $ n \geq 1 $. According to the definition of the symbol $ [x] $,
The products $ na $, $ nb $, $ nc $, $ nd $ cannot be integers (since the numbers $ a $, $ b $, $ c $, $ d $ are irrational by assumption), and therefore in none of the listed relationships can equality hold. Adding the inequalities in the left "column" and the right "column" (and using the assumption that $ a + b = 1 $), we obtain the following relationships
where
The numbers $ n $ and $ k_n $ are integers, so the first of the double inequalities (1) implies the equality $ n = k_n + 1 $.
If now $ c + d = 1 $, then the number $ l_n $, like $ k_n $, must (by the second pair of inequalities (1)) satisfy the equality $ n = l_n + 1 $; thus $ k_n = l_n( = n-1) $ for $ n = 1,2,3,\ldots $.
Conversely, assuming that $ k_n = l_n $ for $ n = 1,2,3,\ldots $, we obtain from the second pair of relationships (1) the double inequality
and since $ k_n = n - 1 $, we have
The only value of the sum $ c + d $ that satisfies this infinite system of double inequalities is the number $ 1 $.
We have thus established the desired equivalence: | proof | Number Theory | proof | Yes | Yes | olympiads | false | 791 |
LIX OM - III - Task 5
The areas of all sections of the parallelepiped $ \mathcal{R} $ by planes passing through the midpoints of three of its edges,
none of which are parallel and do not have common points, are equal. Prove that the parallelepiped
$ \mathcal{R} $ is a rectangular parallelepiped. | Let's assume that the areas of all the sections mentioned in the problem are equal to $S$.
Let $ABCD$ and $EFGH$ be the bases of the parallelepiped $\mathcal{R}$ (Fig. 3), let $O$ be its center of symmetry, and let $I, J, K, L, M, N$ denote the midpoints of the edges $AE, EF, FG, GC, CD, DA$, respectively. Then the following pairs of points: $I$ and $L$, $J$ and $M$, $K$ and $N$, are symmetric with respect to the point $O$.
om59_3r_img_3.jpg
The plane $\pi$ passing through the points $I, K, M$ in the section of the parallelepiped $\mathcal{R}$ defines a figure with area $S$. We will show that this figure is the hexagon $IJKLMN$.
Let $\pi$ be the plane passing through the points $I, N, M$. By the converse of Thales' theorem, we have $NM \parallel AC \parallel IL$. The line $NM$ and the point $I$ lie in the plane $\pi$, so the line passing through the point $I$ and parallel to the line $NM$ (i.e., the line $IL$) also lies in the plane $\pi$. Therefore, the midpoint $O$ of the segment $IL$ lies in the plane $\pi$. This means that the points $J$ and $K$, which are symmetric to the points $M$ and $N$ with respect to the point $O$, also lie in the plane $\pi$. Consequently, the points $I, J, K, L, M, N$ lie in this plane; they thus form a flat hexagon with the center of symmetry $O$.
Moreover, the planes $\pi$ and $\pi$ have three non-collinear points in common: $I, K$, and $M$. This proves that $\pi = \pi$.
In this way, we have shown that the point $O$ is the center of symmetry of the hexagon $IJKLMN$ with area $S$. Therefore, the quadrilateral $ILMN$ has an area of $\frac{1}{2}S$; it is also a trapezoid, as we have shown earlier that the lines $IL$ and $MN$ are parallel.
If we denote by $P$ and $Q$ the midpoints of the edges $AB$ and $BC$, respectively, and conduct similar reasoning, we can justify that the quadrilateral $ILQP$ is a trapezoid with an area of $\frac{1}{2}S$.
Thus, the trapezoids $ILMN$ and $ILQP$ have equal areas; they also have a common base $IL$ and equal second bases ($NM = PQ = \frac{1}{2}AC$). It follows that these trapezoids have equal heights. In other words, the distances from the line $IL$ to the lines $NM$ and $PQ$ are the same.
If $l_1, l_2$ are different parallel lines in space, then any line parallel to them and equally distant from them lies in the plane $\Pi$, with respect to which the lines $l_1$ and $l_2$ are symmetric. The plane $\Pi$ is also perpendicular to the plane defined by the lines $l_1$ and $l_2$.
Therefore, the lines $AC$ and $IL$, which are equidistant from the lines $NM$ and $PQ$, define a plane perpendicular to the plane $MNPQ$. In other words, the plane $\pi_1$ containing the parallelogram $ACGE$ is perpendicular to the base $ABCD$. Similarly, we show that the plane $\pi_2$ containing the parallelogram $BFHD$ is perpendicular to the base $ABCD$. Hence, the edge of the planes $\pi_1$ and $\pi_2$, which is a line parallel to $AE$, is perpendicular to the base $ABCD$.
From this, we obtain the equalities $\measuredangle EAB = \measuredangle EAD = 90^{\circ}$.
Similarly, we prove that $\measuredangle DAB = 90^{\circ}$. Therefore, the parallelepiped $\mathcal{R}$ is a rectangular parallelepiped. | proof | Geometry | proof | Yes | Yes | olympiads | false | 792 |
LVIII OM - I - Problem 12
The polynomial $ W $ with real coefficients takes only positive values in the interval $ \langle a;b\rangle $ (where $ {a<b} $). Prove that there exist polynomials $ P $ and $ Q_1,Q_2,\ldots,Q_m $ such that
for every real number $ x $. | We will conduct the proof by induction on the degree of the polynomial $W$.
If $W(x) \equiv c$ is a constant polynomial, then of course $c > 0$ and in this case we can take $P(x) \equiv \sqrt{c}$, $m=1$, and $Q_1(x) \equiv 0$.
Now suppose that the thesis of the problem is true for all polynomials of degree less than $n$ and let $W$ be a polynomial of degree $n$.
The function
considered in the interval $(a; b)$, is continuous and takes positive values in it, which approach infinity at the ends of the interval. Therefore, the function $f$ attains its minimum value $c$ at some point (not necessarily one) in the interval; this is a positive number.
Thus, the inequality $f(x) \geq c$ holds, i.e., $W(x) \geq c(x-a)(b-x)$ for $x \in [a; b]$, which becomes an equality at some point (points). Speaking figuratively, $c$ is the number for which the parabola $y = c(x-a)(b-x)$ in the interval $[a; b]$ is "tangent from below" to the graph of the polynomial $W$ on this interval, possibly at several points (Fig. 7). Therefore, the polynomial $G(x) = W(x) - c(x-a)(b-x)$ in the considered interval takes non-negative values and has at least one root (Fig. 8).
om58_1r_img_7.jpg
om58_1r_img_8.jpg
Notice that $G(a) = W(a)$ and $G(b) = W(b)$, so the polynomial $G$ takes positive values at the ends of the interval. Therefore, the roots of the polynomial $G$ in the interval $(a, b)$ have even multiplicity. Thus, there exist numbers (not necessarily distinct) $g_1, g_2, \ldots, g_k \in (a, b)$ such that the polynomial $G$ has the form
$$G(x) = (x-g_1)^2(x-g_2)^2 \ldots (x-g_k)^2H(x),$$
where the polynomial $H$ takes only positive values in the interval $[a; b]$. Let $B(x) = (x-g_1)(x-g_2) \ldots (x-g_k)$; thus, we have $G(x) = B(x)^2H(x)$.
If $m$ is the degree of the polynomial $G$, then $m \leq n$, with the exception of the case $n=1$, when $m=2$. Moreover, the polynomial $H$ has a degree no greater than $m-2$. Therefore, $H$ is a polynomial of degree lower than $W$. Hence, by the induction hypothesis, there exists a representation of the form
Thus, we obtain
and this is the desired representation of the polynomial $W$. This completes the solution of the problem. | proof | Algebra | proof | Yes | Yes | olympiads | false | 793 |
XXVII OM - II - Problem 6
Six points are placed on a plane in such a way that any three of them are vertices of a triangle with sides of different lengths. Prove that the shortest side of one of these triangles is also the longest side of another one of them. | Let $P_1, P_2, \ldots, P_6$ be given points. In each of the triangles $P_iP_jP_k$, we paint the shortest side red. In this way, some segments $\overline{P_rP_s}$ are painted red, while others remain unpainted.
It suffices to prove that there exists a triangle with vertices at the given points, all of whose sides are painted red. The longest side of this triangle is also the shortest side of some other triangle, since it has been painted red. From each of the given points, five segments connect it to the other given points. Therefore, either at least 3 of these segments are painted red, or at least 3 are not painted.
If from point $P_1$ at least three segments are painted red (for example, segments $\overline{P_1P_2}$, $\overline{P_1P_3}$, $\overline{P_1P_4}$ are painted red), then in the triangle determined by the other ends of these segments (i.e., in the triangle $P_2P_3P_4$), at least one of the sides (namely, the shortest one) is painted red. Let, for example, segment $\overline{P_2P_3}$ be painted red. Then in the triangle $P_1P_2P_3$, all sides are painted red.
If, however, from point $P_1$ at least three segments are not painted (let these be segments $\overline{P_1P_2}$, $\overline{P_1P_3}$, $\overline{P_1P_4}$), then consider the triangles $P_1P_2P_3$, $P_1P_2P_4$, $P_1P_3P_4$. In each of them, at least one of the sides is painted red, but it is not the side containing vertex $P_1$. Therefore, the segments $\overline{P_2P_3}$, $\overline{P_2P_4}$, $\overline{P_3P_4}$ are painted red, i.e., the triangle $P_2P_3P_4$ has all its sides painted red.
Note 1. In the above solution, we did not use the assumption that the given points lie in a plane.
Note 2. The thesis of the problem can also be obtained by reasoning similarly to the solution of problem 15 (3) from the XXII Mathematical Olympiad. | proof | Geometry | proof | Yes | Yes | olympiads | false | 795 |
XV OM - I - Problem 9
Prove that the quotient of the sum of all natural divisors of an integer $ n > 1 $ by the number of these divisors is greater than $ \sqrt{n} $. | Let $ d_1, d_2, \ldots, d_s $ denote all the natural divisors of a given integer $ n > 1 $. Among these divisors, there are certainly unequal numbers, such as $ 1 $ and $ n $. According to the Cauchy inequality, the arithmetic mean of positive numbers that are not all equal is greater than the geometric mean of these numbers, hence
We will prove that the right-hand side of this inequality is equal to $ \sqrt{n} $. For this purpose, we will distinguish two cases:
1. The number $ n $ is not a square of an integer. In the set of numbers $ d_1, d_2, \ldots, d_s $, for each number $ d_i $ there is a different number $ d_k = \frac{n}{d_i} $, so the number $ s $ is even and the set $ \{d_1,d_2, \ldots,d_s\} $ can be divided into $ \frac{s}{2} $ such pairs, where the product of the numbers in each pair is equal to $ n $. In this case,
2. The number $ n $ is a square of a natural number $ d_r $. For each number $ d_i $ in the set $ \{d_1, d_2, \ldots, d_s\} $ different from $ d_r $, there is a number $ d_k = \frac{n}{d_i} $ different from $ d_i $, so the number $ s - 1 $ is even and the set of numbers $ \{ d_1, \ldots d_{r-1}, d_{r+1} \ldots, d_s\} $ can be divided into $ \frac{s-1}{2} $ pairs with a product equal to $ n $. In this case,
In view of inequality (1), in each of the cases 1 and 2, the inequality holds:
This completes the translation, maintaining the original text's line breaks and formatting. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 796 |
XXXIV OM - I - Problem 7
Let $ S_n $ be the set of sequences of length $ n $ with terms $ -1 $, $ +1 $. We define the function $ f: S_n - \{(-1, 1, 1,\ldots, 1)\} \to S_n $ as follows: if $ (a_1, \ldots, a_n) \in S_n - \{(-1, 1, 1,\ldots, 1)\} $ and $ k = \max_{1\leq j \leq n} \{j \;:\; a_1\cdot a_2\cdot \ldots \cdot a_j = 1\} $, then $ f(a_1,\ldots, a_n) = (a_1, \ldots, a_{k-1}, -a_k, a_{k+1}, \ldots, a_n) $. Prove that | To simplify the notation, we introduce the symbol $ f^{(k)} $ to denote the $ k $-fold iteration of the function $ f $:
The thesis of the problem will be proved by induction. The definition of the function $ f $ depends, of course, on the natural number $ n $. Where it helps to avoid misunderstandings, we will write $ f_n $ instead of $ f $ to highlight the value of the natural number $ n $.
First, we will prove two lemmas.
Lemma 1. If $ m $, $ n $ are natural numbers and $ f_n^{(2m)}(1, 1, \ldots, 1)= (a_1, a_2, \ldots, a_n) $, then $ a_1a_2 \ldots a_n=1 $.
Proof. From the definition of the function $ f $, it follows that if $ f(b_1, b_2, \ldots, b_n)= (c_1, c_2, \ldots, c_n) $, then $ c_1c_2\ldots c_n = -b_1 b_2 \ldots b_n $. Therefore, an even number of compositions of the function $ f $ assigns the sequence $ (1, 1,\ldots, 1) $ a sequence whose terms multiply to $ 1 $.
Lemma 2. If $ f_n^{(m)} (1,1,\ldots,1) = (a_1, a_2, \ldots, a_n) $, then $ f_{n+1}^{(2m)}(1, 1, \ldots,1)= (a_1, a_2, \ldots, a_n, a) $, where $ a = 1 $ or $ a = - 1 $.
Proof. We will use induction on $ m $.
If $ m= 1 $, then $ f_n(1, 1,\ldots, 1)= (1, 1,\ldots, 1, - 1) $, and $ f_{n+1}^{(2)} (1,1,\ldots,1) = f_{n+1}(1,1,\ldots, 1,-1)= (1,1,\ldots,-1,-1) $.
Assume that for some $ m $, if $ f_n^{(m)}(1,1,\ldots, 1) =(a_1,a_2,\ldots,a_n) $, then $ f_{n+1}^{(2m)}(1,1,\ldots, 1) =(a_1, a_2,\ldots,a_n, a) $. From the definition of the function $ f $, it follows that $ f_n^{(m+1)}(1,1,\ldots, 1) =f_n(a_1, a_2, \ldots, a_n)= (a_1, a_2, \ldots, -a_k, \ldots, a_n) $. By Lemma 1, $ a_1a_2\ldots a_na= 1 $, so $ f_{n+1}(a_1, a_2, \ldots, a_n, a)= (a_1, a_2, \ldots, a_n, -a) $, and by setting $ a_{n + 1} = -a $ we have $ f_{n+1}(a_1,a_2,\ldots, a_n,a)= (a_1, a_2, \ldots, a_n, a_{n+1}) $. Since $ a_1a_2\ldots a_{n + 1} = - 1 $, it follows that
$ f_{n+1}(a_1, a_2, \ldots, a_{n+1})= (a_1,a_2,\ldots,-a_k, \ldots, a_{n+1}) $. Therefore, $ f_{n+1}^{(2(k+1))}(1, 1, \ldots, 1)= f_{n+1}^{(2)} (a_1, a_2, \ldots, a_n, -a_{n+1}) = (a_1, a_2, \ldots, -a_k, \ldots, a_n, a_{n+1}) $. This completes the proof of the inductive step. By the principle of induction, the thesis of the lemma is satisfied for every natural $ k $.
We will now proceed to the proof of the theorem stated in the problem.
For $ n =2 $, the theorem is obviously satisfied.
Assume for some $ n $ that
Based on Lemma 2,
where $ a = 1 $ or $ a = -1 $.
However, from Lemma 1, $ -1 \cdot 1 \cdot \ldots \cdot 1 \cdot a = 1 $, so $ a = - 1 $. Therefore,
By the principle of induction, for every $ n $, $ f^{2^{n-1}} (1,1,\ldots, 1)= (-1,1,\ldots, 1) $. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 799 |
XIX OM - I - Problem 7
Points $ D, E, F $ are the midpoints of the sides of triangle $ ABC $. Prove that if the circumcircles of triangles $ ABC $ and $ DEF $ are tangent, then the point of tangency is one of the points $ A $, $ B $, $ C $ and triangle $ ABC $ is a right triangle. | Suppose that the circle $c$ circumscribed around triangle $ABC$ and the circle $k$ circumscribed around triangle $DEF$ are tangent at point $T$ (Fig. 5).
Triangle $DEF$ lies inside circle $c$, so the tangency of circles $c$ and $k$ is internal, and the center $S$ of circle $k$ lies on the ray $TO$. Triangle $DEF$ is similar to triangle $ABC$ in the ratio $1:2$, so the radius $TS$ of circle $k$ is half the radius $TO$ of circle $c$; hence, point $S$ is the midpoint of segment $TO$, with $TO$ being the diameter of circle $k$.
Points $D$, $E$, $F$ of circle $k$ are distinct, so at most one of them can coincide with point $O$. Suppose, for example, that points $D$ and $E$ are different from point $O$.
The chord $BC$ of circle $c$, whose midpoint is point $D$, is perpendicular to the line $OD$, so one of the angles $ODB$ and $ODC$ is inscribed in circle $k$, which means that point $B$ or point $C$ is at the end $T$ of the diameter $OT$ of circle $k$. Similarly, one of the endpoints of the chord $AC$ lies at point $T$. Therefore, vertex $C$ of triangle $ABC$ is at point $T$. In this case, the midpoint $F$ of side $AB$ lies at point $O$. If point $F$ were different from point $O$, then the chord $AB$ of circle $c$ would be perpendicular to the line $OF$, so one of the right angles $OFA$ and $OFB$ would be inscribed in circle $k$, which means that point $A$ or point $B$ would be at point $T$, i.e., would coincide with point $C$, which is impossible.
Angle $C$ of triangle $ABC$ is an inscribed angle in circle $c$ whose sides pass through the endpoints of the diameter $AB$ of this circle, i.e., it is a right angle. | proof | Geometry | proof | Yes | Yes | olympiads | false | 803 |
VIII OM - III - Task 5
Given a line $ m $ and a segment $ AB $ parallel to it. Divide the segment $ AB $ into three equal parts using only a ruler, i.e., by drawing only straight lines. | We first find the midpoint $ S $ of segment $ AB $ as in problem 17 (Fig. 16) (Fig. 23). We draw lines $ AT $ and $ SE $, which intersect at point $ G $, and line $ GB $ intersecting line $ m $ at point $ H $. Since $ AS = SB $ and $ m \parallel AB $, then $ DT = TE $ and $ TE = EH $. Therefore, if $ K $ is the point of intersection of lines $ AD $ and $ BH $, then lines $ KT $ and $ KE $ divide segment $ AB $ into $ 3 $ equal parts. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 804 |
LIV OM - III - Task 1
In an acute triangle $ABC$, segment $CD$ is an altitude. Through the midpoint $M$ of side $AB$, a line is drawn intersecting rays $CA$ and $CB$ at points $K$ and $L$, respectively, such that $CK = CL$. Point $S$ is the center of the circumcircle of triangle $CKL$. Prove that $SD = SM$. | From Menelaus' theorem applied to triangle $ABC$ we get
Let $E$ be the second intersection point of line $CS$ with the circumcircle of triangle $CKL$ (Fig. 1). From the equality $CK = CL$, it follows that $EK = EL$. Moreover, $\measuredangle AKE = 90^\circ = \measuredangle BLE$.
Therefore, the right triangles $AKE$ and $BLE$ are congruent. Hence, we get $AE = BE$, which means $EM \perp AB$, and this gives $CD \parallel EM$.
om54_3r_img_1.jpg
Since $S$ is the midpoint of segment $CE$, its orthogonal projection onto line $AB$ coincides with the midpoint of segment $DM$. Therefore, $SD = SM$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 805 |
XIX OM - I - Problem 2
Let $ p(n) $ denote the number of prime numbers not greater than the natural number $ n $. Prove that if $ n \geq 8 $, then $ p(n) < \frac{n}{2} $ | If $ n $ is an even number, then in the set of natural numbers from $ 1 $ to $ n $, there are $ \frac{n}{2}-1 $ even numbers different from $ 2 $, and thus composite, and at least one odd number, namely $ 1 $, which is not a prime number. Prime numbers belong to the set of the remaining $ \frac{n}{2} $ numbers, so $ p(n) \leq \frac{n}{2} $.
If $ n $ is an odd number at least equal to $ 9 $, then in the set of natural numbers from $ 1 $ to $ n $, there are $ \frac{n-1}{2}-1 $ even numbers different from $ 2 $, and there are at least two odd numbers, namely $ 1 $ and $ 9 $, which are not prime numbers. Prime numbers belong to the set of the remaining $ n - \left(\frac{n-1}{2}-1 \right) - 2 = \frac{n-1}{2} $ numbers, so $ p(n) < \frac{n}{2} $.
Therefore, for every natural number $ n $: | proof | Number Theory | proof | Yes | Yes | olympiads | false | 807 |
XXXVII OM - II - Problem 3
Let S be a sphere circumscribed around a regular tetrahedron with an edge length greater than 1. The sphere $ S $ is represented as the union of four sets. Prove that there exists one of these sets that contains points $ P $, $ Q $, such that the length of the segment $ PQ $ exceeds 1. | We will first prove a lemma that is a one-dimensional version of this theorem.
Lemma. The circle $\omega$ circumscribed around an equilateral triangle with a side length greater than $1$ is represented as the union of three sets: $\omega = U \cup V \cup W$. Then, there exist points $P$, $Q$ in one of the sets $U$, $V$, $W$ such that $|PQ| > 1$.
om37_2r_img_7.jpg
Proof of the lemma. For any two non-antipodal points $X, Y \in \omega$, we will denote by $\breve{XY}$ the closed arc of the circle $\omega$ with endpoints $X$ and $Y$, contained in a semicircle. Suppose that the thesis of the lemma is not true and take any equilateral triangle $ABC$ inscribed in $\omega$. Its vertices must belong to different sets (among $U$, $V$, $W$); we can assume that $A \in U$, $B \in V$, $C \in W$. Then $W \cap \breve{AB} = \emptyset$ (because the distance from any point of the arc $\breve{AB}$ to the point $C$ exceeds $1$). Let
and let $E \in \breve{CA}$, $F \in \breve{CB}$ be points such that $|CE| = a$, $|CF| = b$ (figure 7). The segment $EF$ has a length not greater than $1$; if $|EF| > 1$, then since, according to (1), arbitrarily close to the points $E$ and $F$ there are points of the set $W$, we would find points $P, Q \in W$ such that $|PQ| > 1$ - contrary to our assumption. Therefore, $|EF| \leq 1$, and thus the arc $EF$ is smaller than $1/3$ of the circle. (This is, of course, the arc with endpoints $E$ and $F$ that contains the point $C$, because the other of these arcs contains the points $A$ and $B$). There are therefore points $P, Q, R \in \omega$ lying outside the arc $\breve{EF}$ and being the vertices of an equilateral triangle.
From the definition of the numbers $a$ and $b$ and the points $E$ and $F$, it follows that $W \cap \breve{CA} \subset \breve{CE}$, $W \cap \breve{CB} \subset \breve{CF}$, and since $W \cap \breve{AB} = \emptyset$, we conclude from this that $W \subset \breve{EF}$. Therefore, the points $P$, $Q$, $R$ belong to the union of the sets $U$ and $V$, and thus two of them must belong to one of these sets. This completes the proof of the lemma, because $|PQ| = |QR| = |RP| > 1$.
The thesis of the problem immediately follows from the lemma: we take a regular tetrahedron $ABCD$ inscribed in the sphere $S$ and the circle $\omega$ circumscribed around the triangle $ABC$. The point $D$ belongs to one of the four sets under consideration. If this set contains any point $X \in \omega$, then $|DX| > 1$. Otherwise, the circle $\omega$ is contained in the union of the remaining three sets, and by the lemma, one of these sets contains points $P$, $Q$ such that $|PQ| > 1$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 808 |
X OM - III - Task 6
Given is a triangle in which the sides $ a $, $ b $, $ c $ form an arithmetic progression, and the angles also form an arithmetic progression. Find the ratios of the sides of this triangle. | Suppose that triangle $ABC$ satisfies the conditions of the problem, with $A \leq B \leq C$, and thus $a \leq b \leq c$. In this case
From equation (1) and the equation $A + B + C = \pi$, it follows that $B = \frac{\pi}{3}$; the cosine rule then gives the relationship
Eliminating $b$ from equations (2) and (3) yields
From (2) and (4), it follows that
Therefore, if both the sides and angles of the triangle form an arithmetic progression, the triangle is equilateral. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 809 |
XXII OM - II - Problem 1
In how many ways can $ k $ fields of a chessboard $ n \times n $ ($ k \leq n $) be chosen so that no two of the selected fields lie in the same row or column? | First, one can choose $ k $ rows in which the selected fields will lie. This can be done in $ \displaystyle \binom{n}{k} $ ways. Then, in each of these rows, one must sequentially choose one of $ n $ fields, one of $ n - 1 $ fields lying in the remaining columns, etc. - one of $ n - k + 1 $ fields. Therefore, the total number of ways is $ \displaystyle \binom{n}{k} n(n - 1) \ldots (n - k + 1) $. | \binom{n}{k}n(n-1)\ldots(n-k+1) | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 812 |
XXXIII OM - III - Problem 1
Indicate such a way of arranging $ n $ girls and $ n $ boys around a round table so that the number $ d_n - c_n $ is maximized, where $ d_n $ is the number of girls sitting between two boys, and $ c_n $ is the number of boys sitting between two girls. | Consider any arrangement of $ n $ girls and $ n $ boys. Let's call a group of $ k $ girls (for $ k > 1 $) any sequence of $ k $ girls sitting next to each other, if the first and the $ k $-th girl in this sequence are seated next to a boy at the table. Let $ D_n $ be the number of groups of girls in a given arrangement. Similarly, we define a group of $ k $ boys. Let $ C_n $ be the number of groups of boys. Since every group of girls is adjacent to a group of boys or a single boy, and every group of boys is adjacent to a group of girls or a single girl, we have
\[
D_n = C_n
\]
If $ D_n = 0 $, then each girl is adjacent to two boys, which implies that each boy is adjacent to two girls, so $ C_n = 0 $ and $ d_n - c_n = 0 $. If, however, $ D_n \geq 1 $, then from the equality (*), we get
\[
d_n - c_n = \left[ \frac{n}{2} \right] - 1
\]
On the other hand, $ C_n $ is the number of groups of boys, and each group contains at least two boys, so $ C_n \leq \frac{n}{2} $. Moreover, $ C_n $ is an integer, so $ C_n \leq \left[ \frac{n}{2} \right] $. Therefore,
\[
d_n - c_n \leq \left[ \frac{n}{2} \right] - 1
\]
for any arrangement of $ n $ girls and $ n $ boys.
We will now show an arrangement where $ d_n - c_n = \left[ \frac{n}{2} \right] - 1 $. If $ n = 2k $, we seat two boys, one girl, two boys, one girl, and so on, finally the last two boys and the remaining girls. In this arrangement, $ k - 1 $ girls have boys on both sides, while no boy has two female neighbors.
If $ n = 2k + 1 $, we seat similarly: two boys, one girl, two boys, one girl, and so on, finally the last three boys and the remaining girls. In this situation, $ k - 1 $ girls sit between two boys, and no boy has two female neighbors.
Therefore, the maximum value of $ d_n - c_n $ is $ \left[ \frac{n}{2} \right] - 1 $. | [\frac{n}{2}]-1 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 814 |
IX OM - I - Problem 11
Prove that if two quadrilaterals have the same midpoints of their sides, then they have equal areas. Show the validity of an analogous theorem for convex polygons with any even number of sides. | Let $M$, $N$, $P$, $Q$ be the midpoints of the sides $AB$, $BC$, $CD$, $DA$ of quadrilateral $ABCD$ (Fig. 10). By the theorem on the segment joining the midpoints of two sides of a triangle, in triangle $ABD$ we have
and in triangle $BCD$ we have
thus $MQ = NP$ and $MQ \parallel NP$, so quadrilateral $MNPQ$ is a parallelogram; line $BD$ divides it into parallelograms $QMKL$ and $PNKL$. The area of parallelogram $QMKL$ is half the area of triangle $ABD$, since side $MQ$ and the height of the parallelogram relative to this side are equal to half of side $BD$ and half of the height relative to $BD$ in triangle $ABD$. Similarly, the area of parallelogram $PNKL$ is half the area of triangle $BCD$. Therefore, the area of quadrilateral $ABCD$ is twice the area of parallelogram $MNPQ$.
If, therefore, two quadrilaterals have the same midpoints of sides $M$, $N$, $P$, $Q$, then their areas are equal, and specifically equal to twice the area of parallelogram $MNPQ$.
We will prove a more general theorem, that if two convex polygons with $2n$ sides have the same midpoints of sides, then they have equal areas. We will use complete induction. For $n = 2$ the theorem is true; assume it is true for $n = k - 1$ ($k \geq 3$). Let $A_1A_2 \ldots A_{2k}$ and $B_1B_2 \ldots B_{2k}$ be two convex polygons with $2k$ sides having the same midpoints of sides $M_1, M_2, \ldots M_{2k}$, where $M_i$ is the common midpoint of sides $A_iA_{i+1}$ and $B_iB_{i+1}$. In Fig. 11, polygon $B_1B_2 \ldots B_{2k}$ is not drawn; the reader is asked to imagine it.
Draw diagonals $A_1A_4$ and $B_1B_4$ in these polygons, dividing them into quadrilaterals $A_1A_2A_3A_4$ and $B_1B_2B_3B_4$ and convex polygons $A_1A_4 \ldots A_{2k}$ and $B_1 B_4 \ldots B_{2k}$ with $2k - 2$ sides. The midpoint $N$ of segment $A_1A_4$ is uniquely determined by points $M_1$, $M_2$, $M_3$, as it is the fourth vertex of the parallelogram whose other three consecutive vertices are points $M_1$, $M_2$, $M_3$. This means that the same point $N$ is also the midpoint of diagonal $B_1B_4$ of polygon $B_1B_2\ldots B_{2k}$. This implies that both quadrilaterals $A_1A_2A_3A_4$ and $B_1B_2B_3B_4$, as well as both polygons $A_1A_4\ldots A_{2k}$ and $B_1B_4 \ldots B_{2k}$, have the same midpoints of sides. Therefore, by the previous theorem
and by the induction hypothesis
Adding these equalities side by side, we get
Note 1. The proof of the theorem for quadrilaterals applies to both convex and concave quadrilaterals (see Fig. 10). However, in the proof of the theorem for polygons with an even number of sides greater than 4, we relied on the assumption that the polygons are convex, specifically on the fact that drawing a diagonal through the first and fourth vertices (counted in a certain direction along the perimeter) in a convex polygon with $2k$ sides divides the polygon into a quadrilateral and a polygon with $2k - 2$ sides. The question arises whether the theorem is true for concave polygons. The answer is affirmative, but the proof must be conducted differently, as there are concave polygons in which there is no diagonal that cuts off a quadrilateral, such as the 12-pointed star in Fig. 12. We do not provide this proof, leaving it as an interesting but challenging task for the reader.
Note 2. The theorem is also true for polygons with an odd number of sides. In this case, it is obvious, as two polygons with an odd number of sides having the same midpoints of sides must coincide. See Problems from Mathematical Olympiads, Warsaw 1956, PZWS, problem no. 123. | proof | Geometry | proof | Yes | Yes | olympiads | false | 815 |
XXIV OM - II - Problem 2
In a given square, there are nine points, no three of which are collinear. Prove that three of them are vertices of a triangle with an area not exceeding $ \frac{1}{8} $ of the area of the square. | Let's first prove the
Lemma. If triangle $ABC$ is contained within a certain rectangle, then the area of the triangle is not greater than half the area of that rectangle.
Proof. Let $A$, $B$, $C$ be the projections of the vertices of triangle $ABC$ onto one of the sides of the rectangle $PQRS$ containing this triangle (Fig. 14). Suppose, for example, that $B$ and let $D$ be the point of intersection of line $AC$ with the line perpendicular to $PQ$ passing through point $B$.
Triangle $ABC$ is the sum of triangles $ABD$ and $CBD$ (one of them may be degenerate to a segment if $A$ or $B$). Therefore, $S_{ABC} = S_{ABD} + S_{BCD} = \frac{1}{2} A$.
Since $A$ and $BD \leq QR$, it follows that
We now proceed to solve the problem. We divide the given square by lines parallel to its sides into four congruent squares (or four congruent rectangles) - Fig. 15. The area of such a square (rectangle) is equal to $\frac{1}{4}$ of the area of the given square. Among the nine given points, at least three belong to one of the obtained squares (rectangles). The area of the triangle with vertices at these points is, by the lemma, not greater than $\frac{1}{8}$ of the area of the given square.
Note. In a similar way, it can be shown that the statement of the problem remains true if the numbers $9$ and $\frac{1}{8}$ are replaced by the numbers $2n+1$ and $\frac{1}{2n}$, respectively. It is sufficient to consider the division of the square by lines parallel to its sides into $n$ congruent rectangles. | proof | Geometry | proof | Yes | Yes | olympiads | false | 817 |
LI OM - II - Task 2
The bisector of angle $ BAC $ of triangle $ ABC $ intersects the circumcircle of this triangle at point $ D $ different from $ A $. Points $ K $ and $ L $ are the orthogonal projections of points $ B $ and $ C $, respectively, onto the line $ AD $. Prove that | Let $ o $ be the circumcircle of triangle $ ABC $. From the equality of angles $ BAD $ and $ DAC $, it follows that the lengths of arcs $ BD $ and $ DC $ of circle $ o $ are equal (the length of arc $ XY $ is measured from point $ X $ to point $ Y $ in the counterclockwise direction).
om51_2r_img_1.jpg
om51_2r_img_2.jpg
Let $ E $ be the point symmetric to point $ D $ with respect to the perpendicular bisector of side $ AB $, and let $ N $ be the orthogonal projection of point $ E $ onto line $ AD $ (Fig. 1 and 2). Then $ AD = BE $. Moreover, the lengths of arcs $ DC $ and $ EA $ are equal, which means that $ EN = CL $.
Points $ B $ and $ E $ lie on opposite sides of line $ AD $. Therefore, the length of segment $ BE $ is not less than the sum of the distances from points $ B $ and $ E $ to line $ AD $. In other words, $ BE \geq BK + EN $, which means $ AD \geq BK + CL $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 819 |
XII OM - II - Task 3
Prove that for any angles $ x $, $ y $, $ z $ the following equality holds | We transform the right side of equation (1) by first applying the formula $ 2 \sin \alpha \sin \beta = \cos (\alpha - \beta) - \cos (\alpha + \beta) $, and then the formulas for $ \cos (\alpha + \beta) $ and $ \cos (\alpha - \beta) $:
After substituting in the last expression $ \sin^2 x \sin^2 y = (1 - \cos^2 x) (1 - \cos^2 y) $, and expanding and reducing the brackets, we obtain an expression equal to the left side of equation (1). | proof | Algebra | proof | Yes | Yes | olympiads | false | 820 |
XXVII OM - I - Problem 11
Transmitting and receiving stations are sequentially connected: $ S_0 $ with $ S_1 $, $ S_1 $ with $ S_2 $, ..., $ S_n $ with $ S_{n+1} $, ... Station $ S_0 $ transmits a signal of 1 or -1, station $ S_1 $ receives the same signal with probability $ 1 - \varepsilon $, and the opposite signal with probability $ \varepsilon $. Station $ S_1 $ transmits the received signal to station $ S_2 $ under the same rules, then station $ S_2 $ transmits the received signal to station $ S_3 $, and so on. Let $ p^n(i|j) $ denote the probability that station $ S_n $ receives the signal $ i $ given that station $ S_0 $ transmitted the signal $ j $. Calculate $ \lim_{n\to \infty} p^n(i|j) $. | We have of course $ p^n(1 \mid 1) = p^n(-1 \mid -1) $, $ p^n(1 \mid 1) + p^n(1 \mid -1) = 1 $ and $ p^n(-1 \mid 1) + p^n(-1 \mid -1) = 1 $. Therefore, $ p^n(1 \mid -1) = p^n(-1 \mid 1) $.
From the above equalities, it follows that the limits of the sequences $(p^n(1 \mid 1))$ and $(p^n(-1 \mid -1))$ are equal, and similarly, the limits of the sequences $(p^n(1 \mid -1))$ and $(p^n(-1 \mid 1))$ are equal, and the limits of the sequences $(p^n(1 \mid -1))$ and $(p^n(1 \mid 1))$ sum to $1$. If the sequence $(p^n(1 \mid 1))$ does not have a limit, then neither do the sequences $(p^n(-1 \mid -1))$, $(p^n(1 \mid -1))$, and $(p^n(-1 \mid 1))$. It is therefore sufficient to find the limit of the sequence $(p^n(1 \mid 1))$.
If station $ S_0 $ transmitted the signal $ 1 $ and station $ S_{n+1} $ received the signal $ 1 $, then there are two possibilities:
(a) either station $ S_n $ received the signal $ 1 $ and then station $ S_{n+1} $ received the same signal; the probability of this event is $ p^n(1 \mid 1) \cdot (1 - \varepsilon) $,
(b) or station $ S_n $ received the signal $ -1 $ and then station $ S_{n+1} $ received the opposite signal; the probability of this event is $ p^n(1 \mid -1) \cdot \varepsilon = (1 - p^n(1 \mid 1)) \cdot \varepsilon $.
Therefore,
Denoting $ p^k(1 \mid 1) $ more briefly by $ p_k $ for $ k = 1, 2, \ldots $, we can write the above equality in the form
From (1), it follows that $ \left( p_n - \frac{1}{2} \right) $ is a geometric sequence with a common ratio of $ 1 - 2\varepsilon $. The first term of this sequence is $ p_1 - \frac{1}{2} = p^1(1 \mid 1) - \frac{1}{2} = (1 - \varepsilon) - \frac{1}{2} = \frac{1}{2} - \varepsilon $. Therefore, using the formula for the $ n $-th term of a geometric sequence, we get
From the conditions of the problem, it follows that $ 0 \leq \varepsilon \leq 1 $. From (2), we get that if $ \varepsilon = 0 $, then $ p^n = \frac{1}{2} + \frac{1}{2} = 1 $ and therefore $ \lim_{n \to \infty} p_n = 1 $.
If, however, $ \varepsilon = 1 $, then $ p_n = \frac{1}{2} + \frac{1}{2} (-1)^n $. The terms of the sequence $(p_n)$ are therefore alternately equal to $ 0 $ and $ 1 $. This sequence does not have a limit. If, however, $ 0 < \varepsilon < 1 $, then $ -1 < 1 - 2\varepsilon < 1 $ and therefore $ \lim_{n \to \infty}(1 - 2 \varepsilon)^n = 0 $. It follows from this and (2) that in this case $ \lim_{n \to \infty} p^n = \frac{1}{2} $.
We have thus proved that if $ \varepsilon = 0 $, then
if $ 0 < \varepsilon < 1 $, then
and if $ \varepsilon = 1 $, then none of the sequences $(p^n(i \mid j))$, where $ i, j \in \{-1, 1\} $, have a limit. | Algebra | math-word-problem | Yes | Yes | olympiads | false | 821 | |
XXX OM - II - Task 3
In space, a line $ k $ and a cube with vertex $ M $ and edges $ \overline{MA} $, $ \overline{MB} $, $ \overline{MC} $, each of length 1, are given. Prove that the length of the orthogonal projection of the edge $ MA $ onto the line $ k $ is equal to the area of the orthogonal projection of the square with sides $ MB $ and $ MC $ onto a plane perpendicular to the line $ k $. | Without loss of generality, we can assume that the line $k$ passes through the point $M$ (Fig. 12) and that the plane $\pi$ perpendicular to the line $k$ also contains the point $M$. Let $A$ be the orthogonal projection of the point $A$ onto the line $k$.
om30_2r_img_12.jpg
As is known, if two planes intersect at an angle $\varphi$ and one of them contains a figure with area $S$, then the orthogonal projection of this figure onto the other plane has an area of $S \cdot \cos \varphi$.
In our problem, let the angle between the planes $\pi$ and $MBC$ be $\varphi$. Then the angle between the lines perpendicular to these planes will also be $\varphi$, i.e., $\measuredangle AMA$.
The area of the square with sides $\overline{MB}$ and $\overline{MC}$ is $1$. Therefore, the area of the projection of this square onto the plane $\pi$ will be $\cos \varphi$. On the other hand, we have $MA$. It follows from this the thesis of the problem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 822 |
III OM - I - Task 4
a) Given points $ A $, $ B $, $ C $ not lying on a straight line. Determine three mutually parallel lines passing through points $ A $, $ B $, $ C $, respectively, so that the distances between adjacent parallel lines are equal.
b) Given points $ A $, $ B $, $ C $, $ D $ not lying on a plane. Determine four mutually parallel planes passing through points $ A $, $ B $, $ C $, $ D $, respectively, so that the distances between adjacent parallel planes are equal. | a) Suppose that the lines $a$, $b$, $c$ passing through points $A$, $B$, $C$ respectively and being mutually parallel satisfy the condition of the problem, that is, the distances between adjacent parallel lines are equal. Then the line among $a$, $b$, $c$ that lies between the other two is equidistant from them. Let this line be, for example, line $b$. In this case, points $A$ and $C$ are equidistant from line $b$ and lie on opposite sides of it; therefore, line $b$ intersects segment $AC$ at its midpoint $M$. From the fact that points $A$, $B$, $C$ do not lie on a straight line, it follows that point $M$ is different from point $B$.
Hence the construction: we draw line $b$ through point $B$ and through the midpoint $M$ of segment $AC$, and then we draw through points $A$ and $C$ lines $a$ and $c$ parallel to line $b$ (Fig. 12). The parallel lines $a$, $b$, $c$ determined in this way solve the problem, since points $A$ and $C$, and thus lines $a$ and $c$, are equidistant from line $b$ and lie on opposite sides of this line.
We found the above solution assuming that among the sought lines $a$, $b$, $c$, the line $b$ lies between lines $a$ and $c$; since the "inner" line can equally well be $a$ or $c$, the problem has three solutions.
b) Suppose that the planes $\alpha$, $\beta$, $\gamma$, $\delta$, passing through points $A$, $B$, $C$, $D$ respectively and being mutually parallel, satisfy the condition of the problem, that is, the distances between adjacent planes are equal. Let these planes be in the order $\alpha$, $\beta$, $\gamma$, $\delta$. We mean by this that plane $\beta$ is equidistant from planes $\alpha$ and $\gamma$, and plane $\gamma$ is equidistant from planes $\beta$ and $\delta$.
In this case, points $A$ and $C$ are equidistant from plane $\beta$ and lie on opposite sides of it, so plane $\beta$ passes through the midpoint $M$ of segment $AC$. Similarly, plane $\gamma$ passes through the midpoint $N$ of segment $BD$. From the fact that points $A$, $B$, $C$, $D$ do not lie in a plane, it follows that point $M$ is different from point $B$, and point $N$ is different from point $C$.
From this, we derive the following construction. We connect point $B$ with the midpoint $M$ of segment $AC$, and point $C$ with the midpoint $N$ of segment $BD$ (Figure 13 shows a parallel projection of the figure). Lines $BM$ and $CN$ are skew; if they lay in the same plane, then points $A$, $B$, $C$, $D$ would lie in the same plane, contrary to the assumption. We know from stereometry that through two skew lines $BM$ and $CN$ one can draw two and only two parallel planes $\beta$ and $\gamma$.
To do this, we draw through point $M$ a line $m$ parallel to line $CN$, and through point $N$ a line $n$ parallel to line $BM$; plane $\beta$ is then determined by lines $m$ and $BM$, and plane $\gamma$ by lines $n$ and $CN$.
Finally, we draw through points $A$ and $D$ planes $\alpha$ and $\delta$ parallel to planes $\beta$ and $\gamma$; we can determine them, as indicated in Figure 13, by drawing through each of points $A$ and $D$ lines parallel to lines $BM$ and $CN$.
The planes $\alpha$, $\beta$, $\gamma$, $\delta$ determined in this way solve the problem, since points $A$ and $C$, and thus planes $\alpha$ and $\gamma$, are equidistant from plane $\beta$ and lie on opposite sides of this plane - and similarly, planes $\beta$ and $\delta$ are equidistant from plane $\gamma$ and lie on opposite sides of this plane.
We found the above solution assuming that the sought planes are in the order $\alpha$, $\beta$, $\gamma$, $\delta$. For other orders of the sought planes, we will find other solutions in the same way. The number of all possible orders, or permutations of the letters $\alpha$, $\beta$, $\gamma$, $\delta$, is $4!$, i.e., $24$. However, note that two "reverse" permutations, such as $\alpha$, $\beta$, $\gamma$, $\delta$ and $\delta$, $\gamma$, $\beta$, $\alpha$, give the same solution. Therefore, the problem has $\frac{24}{2} = 12$ solutions corresponding to the permutations. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 823 |
LV OM - I - Task 2
Determine whether there exists a prime number $ p $ and non-negative integers $ x $, $ y $, $ z $ satisfying the equation | Suppose there exist numbers $ p $, $ x $, $ y $, $ z $ satisfying the given equation in the problem. Since $ p $ is a prime number, the divisors of $ p^z $ are only powers of $ p $. Therefore, there exist non-negative integers $ a $, $ b $ such that $ 12x + 5 = p^a $ and $ 12y + 7 = p^b $. From this, it follows that the number $ p $ is odd and not divisible by 3, so it must be of the form $ 6k + 1 $ or $ 6k - 1 $, where $ k $ is a positive integer. Then
it follows that
On the other hand, $ p^a \equiv 5 (\mathrm{mod}\ 12) $ and $ p^b \equiv 7 (\mathrm{mod}\ 12) $. But by the relation (*), both of these congruences cannot hold simultaneously. The contradiction obtained proves that there do not exist numbers satisfying the given conditions. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 826 |
LVII OM - III - Problem 4
We perform the following operation on a triplet of numbers. We select two of these numbers and replace them with their sum and their product, while the remaining number remains unchanged. Determine whether, starting from the triplet (3,4,5) and performing this operation, we can again obtain a triplet of numbers that are the lengths of the sides of a right-angled triangle. | If in the first step we choose the numbers 3 and 5, we will get the triplet (4, 8, 15), in which exactly one number is odd. Choosing two even numbers from such a triplet, we replace them with even numbers, whereas deciding on one even and one odd number, we get numbers of different parities. Therefore, from a triplet in which exactly one number is odd, we will get a triplet that has the same property.
If, however, among the natural numbers $a, b, c$ exactly one is odd, then the equality $a^2 + b^2 = c^2$ cannot hold — one side of this equation is odd, and the other is even. We will not, therefore, obtain a triplet of numbers that are the lengths of the sides of a right-angled triangle.
On the other hand, choosing the numbers 3 and 4 in the first step, we will get the triplet (5, 7, 12), and taking 4 and 5 will yield the triplet (3, 9, 20). In both of these triplets, there is exactly one even number, and it is the largest number in the triplet. If in such a triplet we choose two odd numbers, we will get a triplet in which exactly one number is odd. From such a triplet, as we already know, we will not obtain a triplet of numbers that are the lengths of the sides of a right-angled triangle.
If, however, we consistently choose numbers of different parities, in each step we will get a triplet with exactly one even number, and from the inequality $xy > x + y$ (true for numbers $x, y > 2$), it follows that this even number will be the largest number in the triplet.
It remains to note that if the natural numbers $a, b$ are odd, and the number $c$ is even, then $a^2 + b^2 \equiv 1 + 1 (\mod 4)$,
so the equality $a^2 + b^2 = c^2$ cannot be satisfied.
Answer: Re-obtaining a triplet of numbers that are the lengths of the sides of a right-angled triangle is not possible. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 827 |
XLVI OM - I - Problem 8
In a regular $ n $-sided pyramid, the angles of inclination of the lateral face and the lateral edge to the base plane are $ \alpha $ and $ \beta $, respectively. Prove that $ \sin^2 \alpha - \sin^2 \beta \leq \tan ^2 \frac{\pi}{2n} $. | Let's assume that the side of the regular $ n $-gon $ A_1A_2\ldots A_n $, which serves as the base of the pyramid, has a length of $ 2a $. Let $ O $ be the common center of the circle circumscribed around this polygon and the circle inscribed in it; let the radii of these circles be (respectively) $ R $ and $ r $. Denote the vertex of the pyramid by $ S $, and the length of the segment $ OS $ (the height of the pyramid) by $ h $.
om46_1r_img_4.jpg
Take one of the edges of the base - for example, $ A_1A_2 $ - and let $ M $ be its midpoint. Let $ \varphi = | \measuredangle A_1OM| = | \measuredangle A_2OM| = \pi/n $. The dihedral angle between the plane of the base (i.e., the plane $ A_1A_2O $) and the plane of the lateral face $ A_1 A_2S $ has a measure equal to the planar angle $ OMS $. From the adopted notation, the following relationships (figure 4) follow:
It is necessary to show that
The useful trigonometric relationships are:
Using formulas (4) and (2), we obtain the equality
Based on the relationships (5) and (1), we have:
Therefore,
And since
the numerator of the fraction obtained on the right side of equation (7) is not greater than the denominator. Therefore, the value of the quotient (7) does not exceed $ 1 $; thus, inequality (3) has been proven. | proof | Geometry | proof | Yes | Yes | olympiads | false | 828 |
XXV OM - III - Problem 1
In the tetrahedron $ ABCD $, the edge $ \overline{AB} $ is perpendicular to the edge $ \overline{CD} $ and $ \measuredangle ACB = \measuredangle ADB $. Prove that the plane determined by the edge $ \overline{AB} $ and the midpoint of the edge $ \overline{CD} $ is perpendicular to the edge $ \overline{CD} $. | We will first prove the
Lemma. If lines $AB$ and $PQ$ intersect at a right angle at point $P$, then the number of points on the ray $PQ^\to$ from which the segment $\overline{AB}$ is seen at an angle $\alpha$ is $0$, $1$, or $2$.
Proof. As is known, the set of points contained in the half-plane with edge $AB$ and containing point $Q$, from which the segment $\overline{AB}$ is seen at an angle $\alpha$, is an arc of a certain circle. This arc has $0$, $1$, or $2$ points in common with the ray $PQ^\to$ depending on the position of the segment $AB$ relative to point $P$ and the size of the angle $\alpha$ (Fig. 19).
Corollary. Let line $AB$ be perpendicular to the plane $\pi$. The set of points in the plane $\pi$ from which the segment $\overline{AB}$ is seen at an angle $\alpha (0 < \alpha < \pi)$ is empty, is a circle, or is the union of two concentric circles.
Proof. Let $P$ be the point of intersection of line $AB$ with the plane $\pi$, and let $Q$ be any point in the plane $\pi$ different from $P$. By the lemma, the ray $PQ^\to$ contains at most two points from which the segment $\overline{AB}$ is seen at an angle $\alpha$. When rotated around line $AB$, these points form at most two concentric circles, and from each point of these circles, the segment $\overline{AB}$ is seen at an angle $\alpha$. By the lemma, from each point in the plane $\pi$ not belonging to these circles, the segment $\overline{AB}$ is seen at an angle different from $\alpha$.
We proceed to the solution of the problem. Let $\pi$ be a plane perpendicular to $AB$ and containing the segment $\overline{CD}$, and let $P$ be the point of intersection of line $AB$ with the plane $\pi$. By the conditions of the problem, the segment $\overline{AB}$ is seen from points $C$ and $D$ at the same angle. From the corollary to the lemma, it follows that either
1) points $C$ and $D$ belong to one circle centered at point $P$, or,
2) points $C$ and $D$ belong to different circles centered at point $P$.
In the first case, the perpendicular bisector of the segment $\overline{CD}$ contains point $P$, and therefore the plane determined by the edge $\overline{AB}$ and the midpoint of the edge $\overline{CD}$ is perpendicular to the edge $\overline{CD}$. In the second case, the perpendicular bisector of the segment $\overline{CD}$ does not contain point $P$. If it did, then point $P$ would be equidistant from points $C$ and $D$. These points would then belong to one circle centered at point $P$, which is not the case. Therefore, the plane determined by the edge $\overline{AB}$ and the midpoint of the edge $\overline{CD}$ does not contain the perpendicular bisector of the segment $\overline{CD}$, and thus is not perpendicular to it.
The thesis of the problem is therefore not true in the second case. The first case occurs, for example, when point $P$ lies on the segment $\overline{AB}$, and thus, for example, when the angle $\measuredangle ACB$ is obtuse. | proof | Geometry | proof | Yes | Yes | olympiads | false | 831 |
XXIII OM - I - Problem 6
Determine for which digits $ a $ the decimal representation of a number $ \frac{n(n+1)}{2} $ ($ n\in \mathbb{N} $) consists entirely of the digit $ a $. | Of course, $ a $ cannot be zero. If the decimal representation of the number
$ \displaystyle t_n = \frac{1}{2} n (n + 1) $ consists of $ k $ ones, where $ k \geq 2 $, then $ 9 t_n = 10^k - 1 $. From this, after simple transformations, we obtain $ (3n + 1) \cdot (3n + 2) = 2 \cdot 10^k = 2^{k+1} \cdot 5^k $. Since the consecutive natural numbers $ 3n + 1 $ and $ 3n + 2 $ are relatively prime and $ 2^{k+1} \leq 2^{2k} = 4^k < 5^k $, it follows that $ 3n + 1 = 2^{k+1} $ and $ 3n + 2 = 5^k $. Subtracting these equations side by side, we get $ 1 = 5^k - 2^{k+1} > 4^k - 2^{k+1} = 2^k (2^k - 2) \geq 8 $, since $ k \geq 2 $. The obtained contradiction proves that $ a $ cannot be one.
Notice that $ 8 t_n + 1 = (2n + 1)^2 $. If the decimal representation of the number $ t_n $ ends with the digits $ 2 $, $ 4 $, $ 7 $, or $ 9 $, then the decimal representation of the number $ 8 t_n + 1 $ would end with the digits $ 7 $ or $ 3 $. However, the last digit of the square of any natural number can only be $ 0, 1, 4, 5, 6 $, or $ 9 $. Therefore, $ a \ne 2, 4, 7, 9 $.
If the decimal representation of the number $ t_n $ ends with the digits $ 33 $ or $ 88 $, then the decimal representation of the number $ 8 t_n + 1 $ would end with the digits $ 65 $ or $ 05 $. However, the decimal representation of the square of any natural number $ m $ does not end with the digits $ 65 $ or $ 05 $. Such a number $ m $ would be odd and divisible by $ 5 $, i.e., $ m = 10k + 5 $. Therefore, the number $ m^2 = 100k^2 + 100k + 25 $ would give a remainder of $ 25 $ when divided by $ 100 $, meaning the decimal representation of the number $ m^2 $ would end with the digits $ 25 $. Therefore, $ a \ne 3, 8 $.
However, $ a $ can be equal to $ 5 $ or $ 6 $. We have $ t_{10} = 55 $, $ t_{11} = 66 $, $ r_{36} = 666 $.
Note. It has been proven that among the numbers $ t_n $ for $ n > 3 $, only the numbers $ t_{10} $, $ t_{11} $, and $ t_{36} $ have a decimal representation composed of the same digits. See David W. Ballew and Ronald C. Weger in Notices Amer. Math. Soc. 19 (1972), p. A - 511, abstract 72 T - A 152. | 5,6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 832 |
XXXVIII OM - III - Problem 2
A regular $ n $-gon is inscribed in a circle of length 1. From among the arcs of the circle having endpoints at the vertices of this polygon, we randomly draw five arcs $ L_1, \ldots, L_5 $ with replacement, where the probability of drawing each arc is the same. Prove that the expected value of the length of the common part $ L_1\cap L_2 \cap L_3 \cap L_4 \cap L_5 $ does not depend on $ n $. | The vertices of a polygon divide a circle into $n$ arcs of length $1/n$. Let us number these arcs: $J_1, \ldots, J_n$. Denote by $\mathcal{L}$ the set from which the selection is made. The elements of the set $\mathcal{L}$ are arcs of a given circle, each of which has two ends, which are different vertices of the considered polygon (see: Note). In other words: an arc $L$ of a given circle is an element of the set $\mathcal{L}$ if and only if it is the sum of $m$ arcs $J_i$ where $0 < m < n$.
Let us fix a number $i \in \{1, \ldots, n\}$. Choose a random arc $L \in \mathcal{L}$. The probability that the arc $L$ contains the fixed arc $J_i$ equals $1/2$:
This follows from the fact that each pair of vertices of the polygon defines two different arcs, of which exactly one contains $J_i$, and thus the arcs $K \in \mathcal{L}$ that contain our arc $J_i$ constitute exactly half of the set $\mathcal{L}$.
The experiment considered in the problem involves drawing five arcs $L_1, L_2, L_3, L_4, L_5 \in \mathcal{L}$, with repetitions allowed. We assume that the drawings are independent: the probability that the intersection of the drawn arcs contains the fixed arc $J_i$ equals the product of the probabilities (1), taken sequentially for $L = L_1, \ldots, L_5$:
For $i \in \{1, \ldots, n\}$, let $X_i$ denote the random variable defined as follows:
In view of equation (2), the expected value of this variable is
The intersection $L_1 \cap L_2 \cap L_3 \cap L_4 \cap L_5$ can be an empty set, a one-point set, or a union of some number of arcs $J_i$. Its length - the random variable $X$ we are interested in - equals the number of arcs $J_i$ contained in this union, multiplied by $1/n$ (the common length of all $J_i$). Therefore, $X$ takes the value $k/n$ if and only if exactly $k$ of the random variables $X_1, \ldots, X_n$ have the value $1$, and the rest have the value $0$. Consequently,
\[ X = \frac{1}{2} (X_1 + \ldots + X_n) \]
and as a result, the expected value of the variable $X$ equals the arithmetic mean of the values (3), i.e.,
The obtained number, as can be seen, does not depend on $n$.
Note. The formulation of the problem does not clearly indicate whether the word "ends" used in the plural carries the information that each of the considered arcs $L$ has two different ends - that is, whether "improper" arcs, with coinciding ends, and thus reducing to a single point, or conversely, being the entire circle, are not considered. In response to participants' questions about this, it was stated that "improper" arcs are not taken into account - that is, as in the solution given above. However, it is worth noting that the result and the entire reasoning do not change if we include "improper" arcs in the set $\mathcal{L}$, provided that we assume the existence of $n$ one-point arcs and also arcs obtained by traversing the entire circle (beginning and end at any vertex). Still, each of the arcs in $\mathcal{L}$ will be contained in exactly half of the arcs in $L$, so formula (1) will remain valid.
Let us also note that the assumption of independence of the five consecutive drawings is here taken for granted. | \frac{1}{32} | Combinatorics | proof | Yes | Yes | olympiads | false | 833 |
XXXIX OM - III - Problem 3
Let $ W $ be a polygon (not necessarily convex) with a center of symmetry. Prove that there exists a parallelogram containing $ W $ such that the midpoints of the sides of this parallelogram lie on the boundary of $ W $. | Among all triangles $OAB$, where $A$, $B$ are vertices of the polygon $W$, and $O$ is its center of symmetry, let us choose the triangle with the maximum area (there are at least four such triangles, and there may be more; we choose any one of them). Let $OMN$ be the selected triangle, and let $M$, $N$ be the vertices of the polygon $W$ that are symmetric (respectively) to $M$, $N$ with respect to $O$. Through points $M$ and $M'$, we draw lines $m$ and $m'$ parallel to $NN'$; through points $N$ and $N'$, we draw lines $n$ and $n'$ parallel to $MM'$ (see Figure 10).
om39_3r_img_10.jpg
From the maximality of the area of triangle $OMN$, it follows that no vertex of the polygon $W$ can lie farther from the line $MM'$ than point $N$, or farther from the line $NN'$ than point $M$. Therefore, all vertices lie within the parallelogram $R$ formed by the lines $m$, $m'$, $n$, $n'$, and thus $W \subset R$. The midpoints of the sides of $R$ are the points $M$, $N$, $M'$, $N'$, which are vertices of the polygon. Therefore, $R$ is the desired parallelogram. | proof | Geometry | proof | Yes | Yes | olympiads | false | 834 |
XX OM - II - Task 6
Prove that every polyhedron has at least two faces with the same number of sides. | Let $ s $ denote the number of faces of the polyhedron $ W $, and $ P $ be the face with the largest number of sides, which we will denote by $ n $. The face $ P $ is adjacent along its sides to $ n $ other (pairwise distinct) faces of the polyhedron, so $ s \geq n+1 $. On the other hand, the number of sides of each face of the polyhedron $ W $ is one of the numbers $ 3, 4, \ldots, n $; there are $ n-2 $ such numbers. If each face of the polyhedron had a different number of sides, then the inequality $ s \leq n-2 $ would hold, which contradicts the previous one. Therefore, such a polyhedron does not exist.
Note. The reasoning above demonstrates a stronger theorem: In every polyhedron, there are at least three faces such that each of them has the same number of sides as some other face of the polyhedron. | proof | Geometry | proof | Yes | Yes | olympiads | false | 836 |
XLIX OM - I - Problem 12
Let $ g(k) $ be the greatest prime divisor of the integer $ k $, when $ |k|\geq 2 $, and let $ g(-1) = g(0) = g(1) = 1 $. Determine whether there exists a polynomial $ W $ of positive degree with integer coefficients, for which the set of numbers of the form $ g(W(x)) $ ($ x $ - integer) is finite. | We will prove that there does not exist a polynomial with the given property. For a proof by contradiction, suppose that
is a polynomial of degree $ n \geq 1 $ with integer coefficients and that the set of numbers of the form $ g(W(x)) $ (for integer values of $ x $) is finite. This means: there exists a natural number $ m $ and there are prime numbers $ p_1, \ldots, p_m $ with the following property:
if $ x $ is an integer and $ W(x) \neq Q $, then the value $ W(x) $ is not divisible by any prime number different from $ p_1, \ldots, p_m $.
We will consider two cases.
If $ a_0 = 0 $, then for every integer $ x \neq 0 $, the value $ W(x) $ is divisible by $ x $. Let $ b = 1 + p_1 p_2 \ldots p_m $. We find a natural number $ k $ for which $ W(kb) \neq 0 $; such a number exists because the polynomial $ W $ has only finitely many roots.
The value $ W(kb) $ is divisible by $ kb $, and therefore by $ b $; hence it has at least one prime divisor different from $ p_1, \ldots, p_m $. This is a contradiction to condition (2).
Now we will consider the case when $ a_0 \neq 0 $. Consider the number $ c = a_0 p_1 p_2 \ldots p_m $. We find a natural number $ k \geq 2 $ for which $ W(kc) \neq a_0 $; such a number exists because the value $ a_0 $ is taken by the polynomial $ W $ at only finitely many points.
From formula (1) we obtain the equality
where $ q $ denotes the number in the square brackets. This is an integer different from zero (since $ W(kc) \neq a_0 $). Substituting into (3) the expression defining the number $ c $, we get the equality
Since $ k \geq 2 $ and $ q \neq 0 $, the number $ w $ is not equal to $ 1 $, $ -1 $, or $ 0 $; it therefore has a prime divisor different from $ p_1, \ldots, p_m $. Thus, in this case as well, we have obtained a contradiction to condition (2); the proof is complete. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 837 |
LVII OM - I - Problem 3
An acute triangle $ABC$ is inscribed in a circle with center $O$. Point $D$ is the orthogonal projection of point $C$ onto line $AB$, and points $E$ and $F$ are the orthogonal projections of point $D$ onto lines $AC$ and $BC$, respectively. Prove that the area of quadrilateral $EOFC$ is equal to half the area of triangle $ABC$. | Let $ P $ be the point symmetric to point $ C $ with respect to point $ O $ (Fig. 1). Since triangle $ ABC $ is acute,
points $ A $, $ P $, $ B $, and $ C $ lie on a circle with center $ O $ in this exact order.
om57_1r_img_1.jpg
The areas of triangles $ COE $ and $ POE $ are equal, as these triangles have a common height dropped from vertex $ E $, and segments $ CO $ and $ OP $ are of equal length. Similarly, the areas of triangles $ COF $ and $ POF $ are equal. Therefore, we obtain $ [EOFC] = \frac{1}{2} \cdot [EPFC] $, where the symbol $ [\mathcal{F}] $ denotes the area of figure $ \mathcal{F} $.
To complete the solution, we need to prove that $ [EPFC] = [ABC] $.
Segment $ CP $ is the diameter of the circumcircle of triangle $ ABC $, so lines $ AP $ and $ AC $ are perpendicular. Line $ DE $ is
perpendicular to line $ AC $, so it is parallel to line $ AP $. Therefore, the areas of triangles $ DEA $ and $ DEP $ are equal. Similarly,
the areas of triangles $ DFB $ and $ DFP $ are equal. Thus,
which completes the solution of the problem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 839 |
XL OM - I - Task 4
Prove that it is impossible to cut a square along a finite number of segments and arcs of circles in such a way that the resulting pieces can be assembled into a circle (pieces can be flipped). | Suppose the division mentioned in the task is feasible. The resulting parts of the square are denoted by $K_1, \ldots, K_n$. The boundary of each figure $K_i$ consists of a finite number of line segments and arcs of circles. Among the arcs of circles that are parts of the boundary of $K_i$, some are convex outward from $K_i$, and others are convex inward toward $K_i$. Let the sum of the lengths of the first type of arcs be denoted by $a_i$, and the lengths of the second type by $b_i$. We adopt $d_i = a_i - b_i$.
Consider the sum $s = d_1 + \ldots + d_n$.
The figures $K_i$ result from the division of the square. The boundary of the square consists of straight line segments. Therefore, each arc of a circle that is a division line of the square is simultaneously a subset of the boundary of two figures $K_i$, with its length contributing to the sum $s$ with a positive sign in one of these figures and with a negative sign in the other. Thus, $s = 0$.
We assumed that the figures $K_i$ can be assembled into a circle. Some of the considered arcs must then be used to form the boundary of this circle. When calculating the sum $s$, the lengths of these arcs will be counted once, always with a positive sign. On the other hand, the lengths of the remaining arcs of circles, forming the division lines of the circle, will be reduced, as in the case of the square. The sum $s$ will therefore be equal to the circumference of the circle.
The obtained contradiction ($s = 0$ and simultaneously $s > 0$) completes the proof. | proof | Geometry | proof | Yes | Yes | olympiads | false | 841 |
XXVI - I - Problem 11
A ship is moving at a constant speed along a straight line with an east-west direction. Every $ T $ minutes, the direction of movement is randomly chosen: with probability $ p $, the ship moves in the eastern direction for the next $ T $ minutes, and with probability $ q= 1-p $, it moves in the western direction. At a point outside the line, there is a submarine whose task is to torpedo the ship. The travel time of the torpedo from the point of firing to any point on the ship's track is $ 2T $. The captain of the submarine knows the value of $ p $ and aims to maximize the probability of hitting the ship. How should $ p $ be chosen to minimize the probability of the ship being torpedoed? | Let's assume that a submarine fires a torpedo at the moment when a lottery is taking place on the ship. In the case of firing the torpedo at a different time, the reasoning proceeds similarly. We can also assume, without loss of generality, that $ p \geq q $, i.e., $ \displaystyle p \geq \frac{1}{2} $. After $ T $ minutes from the moment the torpedo is fired, the next lottery takes place. The results of these two lotteries can be as follows: east-east, east-west, west-east, west-west with probabilities $ p^2 $, $ pq $, $ qp $, $ q^2 $, respectively. In the first case, after $ 2T $ minutes from the moment the torpedo is fired, the ship will be located at point $ A $ to the east of the initial point $ O $, in the second and third cases - at point $ O $, and in the last case - at point $ B $ to the west of $ O $. The probability of the event that the ship will be at point $ A $, $ O $, $ B $ at that time is therefore $ p^2 $, $ 2pq $, $ q^2 $, respectively.
Since $ p \geq q $, then $ p^2 \geq q^2 $. The inequality $ p^2 > 2pq $ holds if and only if $ p > 2q $. Therefore, if $ p > 2q $, i.e., $ \displaystyle p > \frac{2}{3} $, then the largest of these three numbers is $ p^2 $ and thus the torpedo should be fired in the direction of point $ A $. If, however, $ 2q > p \geq q $, i.e., $ \frac{2}{3} > p \geq \frac{1}{2} $, then the number $ 2pq $ is the largest and the torpedo should be fired in the direction of point $ O $. For $ \displaystyle p = \frac{2}{3} $, we have $ p^2 = 2pq > q^2 $ and thus the torpedo can be fired in the direction of point $ A $ or $ O $ with the same probability of hitting. The probability of hitting $ f(p) $ is therefore
The function $ f(p) $ is increasing in the interval $ \displaystyle \langle \frac{2}{3};\ 1 \rangle $ and decreasing in the interval $ \displaystyle \langle \frac{1}{2};\ \frac{2}{3} \rangle $. Therefore, the smallest probability of hitting is at $ \displaystyle p = \frac{2}{3} $. It is $ \displaystyle \frac{4}{9} $. | \frac{4}{9} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 842 |
XIII OM - I - Problem 11
Given is a quadrilateral $ABCD$ whose diagonals intersect at a right angle at point $M$. Prove that the $8$ points where perpendiculars drawn from point $M$ to the lines $AB$, $BC$, $CD$, and $DA$ intersect the sides of the quadrilateral lie on a circle. | Let $P$, $Q$, $R$, $S$ denote the feet of the perpendiculars dropped from point $M$ to the sides $AB$, $BC$, $CD$, $DA$ respectively (Fig. 16).
First, we will prove that the points $P$, $Q$, $R$, $S$ lie on a circle by showing that the sum of two opposite angles of quadrilateral $PQRS$ equals $180^\circ$.
Each of the quadrilaterals $MPAS$, $MQBP$, $MRCQ$, and $MSDR$ has two right angles, so the vertices of each of them lie on a circle. According to the theorem of inscribed angles,
Thus,
since $AM \bot DM$.
Similarly,
so,
Adding equations (1) and (2) and considering that $\measuredangle MPS + \measuredangle MPQ = \measuredangle SPQ$ and $\measuredangle MRS + \measuredangle MRQ = \measuredangle SRQ$, we get:
Therefore, the points $P$, $Q$, $R$, $S$ indeed lie on a circle. It remains to prove that the points $P_1$, $Q_1$, $R_1$, $S_1$, where the lines $MP$, $MQ$, $MR$, and $MS$ intersect the sides $CD$, $DA$, $AB$, and $BC$ respectively, also lie on the same circle. It is sufficient to prove this for one of these points, for example, $Q_1$, since the reasoning for each of the others is the same (Fig. 17).
Indeed,
From the equality of angles $PSQ_1$ and $PQQ_1$, it follows that the points $P$, $Q$, $S$, $Q_1$ lie on a circle, c.n.d. | proof | Geometry | proof | Yes | Yes | olympiads | false | 843 |
XV OM - III - Task 4
Prove that if the roots of the equation $ x^3 + ax^2 + bx + c = 0 $, with real coefficients, are real, then the roots of the equation $ 3x^2 + 2ax + b = 0 $ are also real. | The task boils down to showing that the given assumptions imply the inequality
Let $ x_1 $, $ x_2 $, $ x_3 $ denote the roots of equation (2); according to the assumption, they are real numbers.
We know that
Hence
Note. Using elementary knowledge of derivatives, the problem can be solved much more simply. It is enough to notice that the function $ 3x^2 + 2ax + b $ is the derivative of the function $ x^3 + ax^2 + bx + c $ and to apply the theorem that between any two roots of a differentiable function, there lies some root of the derivative of that function, and that a multiple root of a function is also a root of its derivative. | proof | Algebra | proof | Yes | Yes | olympiads | false | 844 |
LVIII OM - III - Task 2
A positive integer will be called white if it is equal to 1
or is the product of an even number of prime numbers (not necessarily distinct).
The remaining positive integers will be called black.
Determine whether there exists a positive integer such that the sum of its white
divisors is equal to the sum of its black divisors. | For an integer $ k>1 $, let us denote
We will prove that for any relatively prime positive integers $ l $, $ m $, the equality
is true.
Indeed, every positive divisor $ d $ of the product $ lm $ has a unique representation in the form $ d=ab $, where $ a $ is a divisor of the number $ l $, and $ b $ is a divisor of the number $ m $. Moreover, $ d $ is a white divisor if and only if the numbers $ a $, $ b $ have the same color.
The sum of all products of the form $ ab $, where $ a $, $ b $ are white divisors of the numbers $ l $, $ m $, respectively, is equal to $ B(l)\cdot B(m) $, while the sum of all such products, in which the divisors $ a $, $ b $ are black, is $ C(l)\cdot C(m) $. Therefore,
Similarly, we justify the equality
From the relations (2) and (3), it follows that
which completes the proof of the relation (1).
Now, suppose that the sum of the white divisors of some integer $ n>1 $ is equal to the sum of its black divisors. We then have $ D(n)=0 $.
Let us factorize the number $ n $ into the product of powers of different prime numbers:
By the formula (1), the equality
holds.
Therefore, there exists a prime number $ p $ and a positive integer $ r $ for which $ D(p^r)=0 $. However, this is impossible: all the white divisors of the number $ p^r $ are the numbers $ 1 $, $ p^2 $, $ p^4 $, $\ldots $, and all the black divisors are the numbers $ p $, $ p^3 $, $ p^5 $, $\ldots $, and consequently, the equality
holds, while the number on the right side of the above equality gives a remainder of $ 1 $ when divided by $ p $, so it is different from zero.
Thus, we have reached a contradiction.
Answer: No such number exists. | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 846 |
XLIV OM - II - Problem 5
On the sides $ BC $, $ CA $, $ AB $ of triangle $ ABC $, points $ D $, $ E $, and $ F $ are chosen respectively such that the incircles of triangles $ AEF $, $ BFD $, and $ CDE $ have radii equal to $ r_1 $. The incircles of triangles $ DEF $ and $ ABC $ have radii $ r_2 $ and $ r $, respectively. Prove that $ r_1 + r_2 = r $. | Let $ P $, $ Q $, $ R $ be the points of tangency of the incircle of triangle $ ABC $ with the sides $ BC $, $ CA $, $ AB $, respectively, and let $ K $ be the point of tangency of the incircle of triangle $ EAF $ with the side $ AF $. The centers of these circles are denoted by $ I $ and $ J $, respectively (see Figure 8).
The system of equations
allows expressing $ |AR| $, $ |BP| $, $ |CQ| $ in terms of $ |BC| $, $ |CA| $, $ |AB| $; in particular,
Considering triangle $ EAF $, we have analogously,
Triangles $ ARI $ and $ AKJ $ are similar with a ratio of $ r:r_1 $. Therefore, $ r \cdot |AK| = r_1 \cdot |AR| $, and by the previous equations,
Similarly,
Adding the last three equations, we obtain the relationship,
Denoting half the perimeter of triangle $ ABC $ by $ p $ and half the perimeter of triangle $ DEF $ by $ q $, we rewrite this relationship as,
The area of a triangle is the product of half the perimeter and the length of the radius of the inscribed circle. Therefore, the following relationships hold:
and
The three triangles appearing in equations (2) have pairwise disjoint interiors and fill the triangle $ ABC $ with the triangle $ DEF $ removed. The sum of their areas is thus equal to the difference in the areas of triangles $ ABC $ and $ DEF $, i.e., the difference $ rp - r_2q $. Therefore, adding the equations (2) side by side, we obtain the relationship,
which simplifies to $ rp - r_2q = r_1p + r_1q $, or equivalently,
According to equation (1), the left side of the last equation is equal to $ rq $. Hence, $ r = r_1 + r_2 $. | r_1+r_2 | Geometry | proof | Yes | Yes | olympiads | false | 847 |
IV OM - II - Task 3
A triangular piece of sheet metal weighs $ 900 $ g. Prove that by cutting this sheet along a straight line passing through the center of gravity of the triangle, it is impossible to cut off a piece weighing less than $ 400 $ g. | We assume that the sheet is of uniform thickness everywhere; the weight of a piece of sheet is then proportional to the area of the plane figure represented by that piece of sheet. The task is to show that after cutting a triangle along a straight line passing through its center of gravity, i.e., through the point of intersection of its medians, each part of the triangle has an area of at least $ \frac{4}{9} $ of the area of the entire triangle.
Let $ P $ be the center of gravity of triangle $ ABC $ with area $ S $ (Fig. 36). When the triangle is cut along one of its medians, for example, along $ AD $, it is divided into two triangles with areas equal to $ \frac{1}{2}S $. By cutting the triangle $ ABC $ along a segment passing through point $ P $ and parallel to one of the sides of the triangle, for example, along the segment $ EF $ parallel to side $ AB $, we divide this triangle into triangle $ EFC $ and trapezoid $ ABFE $. Triangle $ EFC $ is similar to triangle $ ABC $ in the ratio $ \frac{EC}{AC} = \frac{2}{3} $. Since the ratio of the areas of similar figures equals the square of the similarity ratio, the area of triangle $ EFG $ equals $ \frac{4}{9} S $, and the area of trapezoid $ ABFE $ equals $ \frac{5}{9}S $.
It turns out that in the two cases considered, the thesis of the theorem holds. Let us draw through point $ P $ any line not passing through any of the vertices of the triangle and not parallel to any of its sides. This line will intersect two sides of the triangle, for example, side $ AC $ at point $ M $ and side $ BC $ at point $ N $. Points $ M $ and $ N $ lie on opposite sides of line $ EF $: for example, let point $ M $ lie on segment $ AE $, and point $ N $ on segment $ FD $. Our theorem will be proved when we show that
To this end, let us note that
Let $ K $ be the point symmetric to point $ A $ with respect to point $ P $. Point $ K $ lies on the extension of segment $ PD $ beyond point $ D $, since $ PK = AP = 2PD $. Triangle $ PFK $ is symmetric to triangle $ PEA $ with respect to point $ P $. Let $ L $ be the point on segment $ FK $ symmetric to point $ M $. Then
From the equality (1) and the inequality (2), it follows that
which was to be proved. | proof | Geometry | proof | Yes | Yes | olympiads | false | 850 |
XII OM - III - Problem 5
Four lines intersecting at six points form four triangles. Prove that the circumcircles of these triangles have a common point. | Four lines intersecting at $6$ points form a figure known as a complete quadrilateral (see Problems from Mathematical Olympiads, vol. I, Warsaw 1960. Problem 79). Let us denote these points by $A$, $B$, $C$, $D$, $E$, $F$ (Fig. 21) in such a way that on the given lines lie the respective triplets of points $(A, B, C)$, $(A, D, E)$, $(C, F, D)$, $(B, F, E)$, with point $B$ lying inside segment $AC$, point $D$ - inside segment $AE$, and point $F$ being a common internal point of segments $CD$ and $BE$.
On this figure, there are triangles: $ABE$, $BCF$, $ACD$, and $DEF$; let $K_1$, $K_2$, $K_3$, $K_4$ denote the circumcircles of these triangles, respectively. Circle $K_2$ passes through point $F$ lying inside circle $K_1$ and through point $C$ lying outside this circle. Therefore, circles $K_1$ and $K_2$ intersect at $2$ points; one of them is point $B$, so the second intersection point $M$ lies on the opposite side of chord $FC$. According to the inscribed angle theorem,
therefore,
Since points $D$ and $M$ lie on the same side of line $AC$, it follows from the equality $\measuredangle AMC = \measuredangle ADC$ that point $M$ lies on circle $K_3$. Circles $K_1$, $K_2$, $K_3$ thus have a common point $M$.
In the same way, we can prove that circles $K_1$, $K_3$, $K_4$ have a common point $N$. Points $M$ and $N$ are common points of circles $K_1$ and $K_3$; neither of them can coincide with the common point $A$ of these circles, since point $A$ lies outside circles $K_2$ and $K_4$, so $M$ and $N$ coincide, i.e., circles $K_1$, $K_2$, $K_3$, $K_4$ have a common point.
Note. The above proof requires supplementation. We have assumed without justification that the intersection points of the $4$ lines can always be labeled in such a way that on the given lines lie the respective triplets $(A, B, C)$, $(A, D, E)$, $(C, F, D)$, $(B, F, E)$, with points $B$, $D$, $F$, and $F$ being internal points of these $4$ triplets. We inferred this from observing the diagram, which is not a rigorous mathematical proof. We will now provide a proof without using the diagram.
Among the given $6$ points, there are $4$ triplets of collinear points, i.e., lying on the same line, and in each triplet, one of the points lies between the other two. Therefore, one of the given $6$ points is not an internal point of any triplet; let us call it $P_1$. Through $P_1$ pass $2$ of the given lines; on one of them lie $2$ more of the given points, which we can label as $P_2$ and $P_3$ in such a way that point $P_2$ lies between points $P_1$ and $P_3$; similarly, on the other line lie $2$ other points, which we will label as $P_4$ and $P_5$ such that $P_4$ lies between $P_1$ and $P_5$. The sixth point we will label as $P_6$, and there can be $2$ cases, which we will consider in turn: 1) $P_6$ is the intersection point of lines $P_3P_4$ and $P_2P_5$; in this case, $P_6$ lies between points $P_3$ and $P_4$, since line $P_2P_5$ intersects side $P_1P_3$ of triangle $P_1P_3P_4$ at point $P_2$, so it must intersect side $P_3P_4$ (Theorem: If a line has a point in common with a side of a triangle and does not pass through any vertex of the triangle, then it has a point in common with one of the other sides of the triangle, is a fundamental fact in geometry usually taken as an axiom known as Pasch's axiom. It was formulated in 1882 by Moritz Pasch (1843-1930), a German mathematician and professor at the University of Giessen.); similarly, point $P_6$ lies between points $P_2$ and $P_5$. If we label points $P_1$, $P_2$, $P_3$, $P_4$, $P_5$, $P_6$ as $A$, $B$, $C$, $D$, $E$, $F$, respectively, we obtain the desired labeling of the points. 2) $P_6$ is the intersection point of lines $P_2P_4$ and $P_3P_5$. In this case, one of two scenarios may occur: a) Point $P_2$ lies between points $P_4$ and $P_6$; in this case, point $P_3$ lies between points $P_5$ and $P_6$, since line $P_1P_3$ passing through point $P_2$ of side $P_4P_6$ of triangle $P_4P_5P_6$ must intersect side $P_5P_6$. We will then obtain the desired labeling by calling points $P_1$, $P_2$, $P_3$, $P_4$, $P_5$, $P_6$ $C$, $F$, $D$, $B$, $A$, $E$, respectively; b) point $P_4$ lies between points $P_2$ and $P_6$; in this case, point $P_5$ lies between points $P_3$ and $P_6$, since line $P_1P_5$ passing through point $P_4$ of side $P_2P_6$ of triangle $P_2P_3P_6$ must intersect some point of side $P_3P_6$. In this case, we label points $P_1$, $P_2$, $P_3$, $P_4$, $P_5$, $P_6$ as $C$, $B$, $A$, $F$, $D$, $E$, respectively, and obtain the desired labeling. | proof | Geometry | proof | Yes | Yes | olympiads | false | 851 |
XXX OM - I - Task 7
For a given natural number $ n $, calculate the number of integers $ x $ in the interval $ [1, n] $ for which $ x^3 - x $ is divisible by $ n $. | If $ n = p^k $, where $ k \geq 1 $ and $ p $ is an odd prime, then for any integer $ x $, at most one of the numbers $ x - 1, x, x + 1 $ is divisible by $ p $. Therefore, the number $ x^3 - x = (x - 1) x(x + 1) $ is divisible by $ p^k $ if and only if one of the numbers $ x - 1, x, x + 1 $ is divisible by $ p^k $, which means that the number $ x $ gives one of the remainders $ 1 $, $ 0 $, or $ -1 $ when divided by $ p^k $.
If $ n = 2^k $, where $ k \geq 3 $, and $ x $ is an even number, then the numbers $ x - 1 $ and $ x + 1 $ are odd. Therefore, the number $ x^3 - x = (x - 1)x(x+1) $ is divisible by $ 2^k $ for even $ x $ if and only if $ x $ is divisible by $ 2^k $, that is, if $ x $ gives the remainder $ 0 $ when divided by $ 2^k $.
If $ n = 2k $, where $ k \geq 3 $, and $ x $ is an odd number, then only one of the consecutive even numbers $ x - 1 $ and $ x + 1 $ is divisible by $ 4 $. Since $ k \geq 3 $, the number $ x^3 - x = (x - 1)x(x + 1) $ is divisible by $ 2^k $ for odd $ x $ if and only if one of the numbers $ x - 1 $ and $ x + 1 $ is divisible by $ 2^{k-1} $, and the other by $ 2 $. This happens when the number $ x $ gives one of the remainders $ 1 $, $ -1 $, $ 2^{k-1} + 1 $, $ 2^{k-1} - 1 $ when divided by $ 2^k $.
Finally, if $ n = 1 $ or $ 2 $, then for any number $ x $ the divisibility $ n \mid x^3 - x $ holds, and if $ n = 4 $, the number is divisible by $ 4 $ if and only if $ x $ gives the remainder $ 1 $, $ 0 $, or $ -1 $ when divided by $ 4 $.
For the general case, we will use the Chinese Remainder Theorem: {\it If $ n_1, n_2, \ldots,n_r $ are pairwise coprime natural numbers, $ n = n_1 \cdot n_2 \ldots n_r $, and $ a_1, a_2, \ldots, a_r $ are any integers, then there exists exactly one integer $ a $ in the interval $ [1, n] $ such that $ n_i \mid a - a_i $ for $ i = 1, 2, \ldots, r $, which means that the numbers $ a $ and $ a_i $ give the same remainder when divided by $ n_i $ for $ i = 1, 2, \ldots, r $.}
Let $ n = 2^k p_1^{k_1} p_2^{k_2} \ldots p_s^{k_s} $, where $ s, k, k_1, k_2, \ldots, k_s $ are integers, $ s \geq 0 $, $ k \geq 0 $, $ k_i > 0 $ for $ i = 1,2, \ldots, s $, and $ p_1, p_2, \ldots, p_s $ are distinct odd primes.
The number $ x^3 - x = (x - 1)x(x + 1) $ is divisible by $ n $ if and only if it is divisible by each of the numbers $ 2^k, p_1^{k_1}, p_2^{k_2}, \ldots, p_s^{k_s} $. By the initial part of the solution, this condition is equivalent to the number $ x $ giving one of the remainders $ 0, 1, -1 $ when divided by $ p_i^{k_i} $ for $ i = 1, 2, \ldots, s $, and
$ x $ gives the remainder $ 0 $ or $ 1 $ when divided by $ 2 $, if $ k = 1 $,
$ x $ gives the remainder $ 0 $, $ 1 $, or $ -1 $ when divided by $ 4 $, if $ k = 2 $,
$ x $ gives one of the remainders $ 0, 1, -1, 2^{k-1} -1, 2^{k-1} + 1 $ when divided by $ 2^k $, if $ k \geq 3 $.
By the Chinese Remainder Theorem, the number of integers $ x $ in the interval $ [1, n] $ that satisfy the conditions of the problem is therefore $ 3^s, 2 \cdot 3^s, 3 \cdot 3^s $, or $ 5 \cdot 3^s $, when $ k = 0 $, $ k = 1 $, $ k = 2 $, or $ k \geq 3 $, respectively. | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 853 | |
XLIII OM - II - Problem 1
Each vertex of a certain polygon has both coordinates as integers; the length of each side of this polygon is a natural number. Prove that the perimeter of the polygon is an even number. | Let $ A_1,\ldots,A_n $ be the consecutive vertices of a polygon; it will be convenient for us to denote the vertex $ A_n $ also by $ A_0 $. The boundary of the polygon forms a closed broken line, hence
Let us denote the coordinates of the $ i $-th vector of the above sum by $ u_i $, $ v_i $, and its length by $ w_i $:
The numbers $ u_i $, $ v_i $, $ w_i $ are assumed to be integers. Of course, .
In the solution, we will use the simple observation that the square of any integer $ m $ is a number of the same parity as $ m $ (since $ m^2 = m + m(m- 1) $, and the product $ m(m- 1) $ is an even number). Hence, for any integers $ m_1 ,m_2,\ldots, m_n $, we have the equivalence:
(Indeed: by the previous observation, both of these sums have the same number of odd terms.)
Equality (1) is equivalent to the statement that
By taking $ m_i= u_i $ and then $ m_i = v_i $ in (2), we conclude that the sums
$ \displaystyle \sum_{i=1}^n u_i^2 $ and $ \displaystyle \sum_{i=1}^n v_i^2 $ are even numbers. Therefore, the sum
is also an even number. Finally, applying equivalence (2) to the numbers $ m_i = w_i $, we infer that the sum $ \displaystyle \sum_{i=1}^n v_i^2 $ is also an even number. And this sum — is precisely the perimeter of the polygon. | proof | Geometry | proof | Yes | Yes | olympiads | false | 854 |
XLII OM - III - Problem 2
Let $ X $ be the set of points in the plane $ (x.y) $ with both coordinates being integers. A path of length $ n $ is any sequence $ (P_0, P_1,\ldots, P_n) $ of points in the set $ X $ such that $ |P_{i-1}P_{i}|=1 $ for $ i \in \{1,2,\ldots,n\} $. Let $ F(n) $ be the number of different paths $ (P_0,P_1,\ldots,P_n) $ starting at $ P_0 = (0,0) $ and ending at $ P_n $ located on the line with the equation $ y = 0 $. Prove that | For any path of length $ n $, with the starting point $ P_0 = (0,0) $, we associate a sequence of $ 2n $ symbols, each of which is a zero or a one. We do this as follows. If the path has the form $ (P_0,P_1,\ldots,P_n) $, then the corresponding zero-one sequence
called the code of the path, is determined by the set of rules:
Speaking figuratively: a step to the right is encoded using the pair $ 10 $; a step up: $ 11 $; a step to the left: $ 01 $; a step down: $ 00 $.
Conversely: every zero-one sequence of length $ 2n $ determines a path of which it is the code.
Different paths correspond to different zero-one sequences; the coding is one-to-one.
Notice now that if the point $ P_n $, the end of the path, lies on the line with the equation $ y = k $, this means that we have taken $ g $ steps up and $ d $ steps down, where $ g $ and $ d $ are numbers with the difference $ g - d = k $. The remaining moves (in the number $ n-g-d $) were horizontal, and it does not matter which of them we made to the right and which to the left.
The number of ones in the code of such a path is
Therefore, the codes of paths ending on the line $ y = 0 $ are sequences composed of exactly $ n $ ones and $ n $ zeros. The positions of the ones can form any $ n $-element subset of the set $ \{1,2,\ldots,2n\} $. The number of such subsets is $ \binom{2n}{n} $. Thus, there are also $ \binom{2n}{n} $ paths of length $ n $ with the end located on the line $ y = 0 $. | \binom{2n}{n} | Combinatorics | proof | Yes | Yes | olympiads | false | 855 |
XXIX OM - I - Problem 4
Let $ Y $ be a figure consisting of closed segments $ \overline{OA} $, $ \overline{OB} $, $ \overline{OC} $, where point O lies inside the triangle $ ABC $. Prove that in no square can one place infinitely many mutually disjoint isometric images of the figure $ Y $. | Let $ d $ be the smallest of the numbers $ OA $, $ OB $, $ OC $. Choose points $ P \in \overline{OA} $, $ Q \in \overline{OB} $, $ R \in \overline{OC} $ such that $ OP= OQ = OR= \frac{d}{2} $ (Fig. 6).
Then $ PQ < OP + OQ = d $ and similarly $ PR < d $ and $ QR < d $. If $ Y $ is a figure consisting of closed segments $ \overline{OA} $, $ \overline{OB} $, and $ \overline{OC} $, and is isometric to the figure $ Y $, and point $ O $ belongs to the triangle $ OPQ $, then each of the segments $ \overline{OA} $, $ \overline{OB} $, and $ \overline{OC} $ intersects the boundary of the triangle $ OPQ $, because in the triangle no segment longer than the longest side of the triangle is contained. From the conditions of the problem, it follows that each of the angles $ \measuredangle A $, $ \measuredangle B $, and $ \measuredangle C $ has a measure less than $ \pi $ and together they form a full angle. Therefore, at least one of the segments $ \overline{OA} $, $ \overline{OB} $, and $ \overline{OC} $ intersects the side $ \overline{OP} $ or $ \overline{OQ} $. The figures $ Y $ and $ Y $ are therefore not disjoint.
om29_1r_img_6.jpg
Similarly, we prove that if point $ O $ belongs to the triangle $ OPR $ or $ OQR $, then the figures $ Y $ and $ Y $ are not disjoint. If, therefore, the figures $ Y $ and $ Y $ are disjoint, then point $ O $ does not belong to the triangle $ PQR $. Let $ r $ be the length of the radius of a circle centered at point $ O $ contained in the triangle $ PQR $. From the above, it follows that if the figures $ Y $ and $ Y $ are disjoint, then the points $ O $ and $ O $ are more than $ r $ apart.
Now consider a square $ K $ with side length $ a $. Choose a natural number $ n $ greater than $ \displaystyle \frac{a}{r} \sqrt{2} $ and divide the square $ K $ by lines parallel to the sides into $ n^2 $ small squares. The length of the side of such a small square is therefore $ \displaystyle \frac{a}{n} $, and the length of its diagonal is $ \displaystyle \frac{a}{n} \sqrt{2} $. Since $ n > \frac{a}{r} \sqrt{2} $, then $ \frac{a}{n} \sqrt{2} < r $. The distance between any two points belonging to a square does not exceed the length of its diagonal. Therefore, the distance between any two points belonging to a small square is less than $ r $. From the initial part of the solution, it follows that if $ f_1 $ and $ f_2 $ are isometries of the figure $ Y $ into the square $ K $ and the figures $ f_1(Y) $ and $ f_2(Y) $ are disjoint, then the points $ f_1(0) $ and $ f_2(0) $ are more than $ r $ apart. These points do not therefore belong to the same small square. Thus, in the square $ K $, one can place at most $ n^2 $ mutually disjoint isometric images of the figure $ Y $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 858 |
XLV OM - I - Problem 5
Prove that if the polynomial $ x^3 + ax^2 + bx + c $ has three distinct real roots, then the polynomial $ x^3 + ax^2 + \frac{1}{4}(a^2 + b)x +\frac{1}{8}(ab - c) $ also has three distinct real roots. | Let's denote the polynomials in the problem by $P(x)$ and $Q(x)$:
Replacing the variable $x$ in the polynomial $P(x)$ with the difference $2x - a$ and transforming the obtained expression:
From the obtained identity:
it immediately follows that if a number $\xi$ is a root of the polynomial $P(x)$, then the number $-\frac{1}{2}(\xi + a)$ is a root of the polynomial $Q(x)$.
Therefore, if the numbers $x_1$, $x_2$, $x_3$ are three distinct roots of the polynomial $P(x)$, then the numbers $-\frac{1}{2}(x_1 + a)$, $-\frac{1}{2}(x_2 + a)$, $-\frac{1}{2}(x_3 + a)$ are three distinct roots of the polynomial $Q(x)$.
{\kom Note.} The idea to transform the expression $P(2x - a)$ does not require any kind of "revelation." The form of the polynomials $P(x)$ and $Q(x)$ (and the thesis of the problem \ldots) suggests that there might be constants $\alpha$, $\beta$, $\gamma$ for which the equality
is satisfied identically (with $\alpha \neq 0$, $\gamma \neq 0$). By comparing the coefficients of $x^3$, we see that if such constants exist, then $\gamma$ must be equal to $\alpha^3$. The proposed identity (2) takes the form
Substituting the expressions defining the polynomials $P(x)$ and $Q(x)$, and equating the coefficients of the polynomials obtained on the left and right sides of formula (2), we obtain a system of three equations with two unknowns ($\alpha$ and $\beta$). It turns out that, regardless of the values of the given constants (parameters) $a$, $b$, $c$, this system always has a solution $\alpha = -2$, $\beta = -a$ (so $\gamma = -8$); in general, this is its only solution, but for some special values of the constants $a$, $b$, $c$, there may be other solutions. However, this does not matter: for the found values of the numbers $\alpha$, $\beta$, $\gamma$, the identity (2) is always satisfied; substituting $x$ with $-x$ it takes the form (1).
Of course, when writing the solution to the problem, no one is required to "explain" where the idea to examine the expression $P(2x-a)$ comes from. | proof | Algebra | proof | Yes | Yes | olympiads | false | 859 |
LI OM - I - Task 5
Determine all pairs $ (a,b) $ of natural numbers for which the numbers $ a^3 + 6ab + 1 $ and $ b^3 + 6ab + 1 $ are cubes of natural numbers. | Let $ a $ and $ b $ be numbers satisfying the conditions of the problem. Without loss of generality, we can assume that $ a \leq b $. Then
Since the number $ b^3+6ab + 1 $ is a cube of an integer, the following equality must hold
Transforming the above relationship equivalently, we obtain step by step
It remains to determine for which pairs $ (a, b) $ satisfying condition (2) the number
is a cube of an integer. By virtue of relation (2), the above number is equal to $ a^3 + 12a^2 - 6a + 1 $. From the inequality
it follows that if the number $ a^3 + 12a^2 - 6a + 1 $ is a cube of an integer, then it is equal to $ (a+ 1)^3 $, $ (a + 2)^3 $, or $ (a + 3)^3 $. Therefore, we need to consider three cases:
(a) $ a^3 + 12a^2 - 6a + 1 = (a + 1)^3 $
In this case, we obtain $ 9a^2 - 9a = 0 $, from which $ a = 0 $ or $ a = 1 $.
(b) $ a^3 + 12a^2 - 6a + 1 = (a + 2)^3 $
Then we have $ 6a^2 - 18a - 7 = 0 $. The obtained equation has no integer solutions, as the left side is not divisible by $ 3 $, while the right side is.
(c) $ a^3 + 12a^2 - 6a + 1 = (a + 3)^3 $
Then we get $ 3a^2 - 33a - 26 = 0 $. Similarly to case (b), the obtained equation has no integer solutions, as the left side is not divisible by $ 3 $, while the right side is equal to $ 0 $.
Ultimately, we have found only one pair $ (a, b) = (1, 1) $ satisfying the conditions of the problem. | (1,1) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 861 |
IX OM - III - Task 1
Prove that the product of three consecutive natural numbers, of which the middle one is a cube of a natural number, is divisible by $ 504 $. | We need to prove that if $a$ is a natural number greater than $1$, then the number
is divisible by $504 = 7 \cdot 8 \cdot 9$. Since the numbers $7$, $8$, and $9$ are pairwise coprime, the task reduces to proving the divisibility of $N$ by each of these numbers.
a) The number $a$ can be represented in the form $a = 7k + r$, where $k$ is a natural number, and $r$ is one of the numbers $0$, $1$, $2$, $3$, $4$, $5$, $6$. Then $a^3 = 7^3k^3 + 3 \cdot 7^2k^2r + 3 \cdot 7kr^2 + r^3$, so when divided by $7$, the number $a^3$ gives the same remainder as $r^3$. But $r^3$ is one of the numbers $0$, $1$, $8$, $27$, $64$, $125$, $216$, so the remainder of $a^3$ when divided by $7$ is one of the numbers $0$, $1$, $6$, which means that one of the numbers $a^3$, $a^3 - 1$, $a^3 + 1$ is divisible by $7$.
b) To establish the divisibility of the number $N$ by $8$, it suffices to note that if $a$ is an even number, then $a^3$ is divisible by $8$, and if $a$ is odd, then $a^3 - 1$ and $a^3 + 1$ are two consecutive even numbers, one of which is divisible by $4$, and their product by $8$.
c) The number $a$ can be represented in the form $a = 3l + s$, where $l$ is a natural number, and $s$ is one of the numbers $0$, $1$, $2$. Then $a^3 = 3^3l^3 + 3 \cdot 3l^2s + 3 \cdot 3l \cdot s^2 + s^3$, from which we see that $a^3$ gives a remainder of $s^3$ when divided by $9$, i.e., $0$, $1$, or $8$. Therefore, one of the numbers $a^3$, $a^3 - 1$, or $a^3 + 1$ is divisible by $9$.
Note 1. It is easy to observe that in the statement of the theorem, we can speak more generally about integers instead of natural numbers.
Note 2. Using simple facts from number theory, the arguments presented above in points a) and c) can be replaced with simpler ones. According to Fermat's Little Theorem (see Seventh Mathematical Olympiad, Warsaw 1957, problem no. 2), if an integer $a$ is not divisible by $7$, then the congruence
holds, from which it follows that
Thus, if $a$ is any integer, one of the numbers $a^3$, $a^3 - 1$, $a^3 + 1$ is divisible by $7$.
When it comes to divisibility by $9$, which is not a prime number, a more general theorem than Fermat's must be applied, namely Euler's Theorem (the famous Swiss mathematician, 1707-1783). Let $m$ be a natural number, and let $\varphi(m)$ denote the number of natural numbers not greater than $m$ and coprime with $m$. For example, $\varphi(1) = \varphi(2) = 1$, $\varphi(3) = \varphi(4) = 2$, $\varphi(5) = 4$, $\varphi(6) = 2$, $\varphi(7) = 6$, $\varphi(8) = 4$, $\varphi(9) = 6$, etc.
Euler's Theorem states: If the integers $a$ and $m > 1$ are coprime, then
For example, if $m = 9$, we get the theorem: if the integer $a$ is not divisible by $3$, then
from which - as above - it follows that for any integer $a$, one of the numbers $a^3$, $a^3 - 1$, $a^3 + 1$ is divisible by $9$.
A proof of Euler's Theorem can be conducted as follows. Let $r_1, r_2, \ldots, r_\varphi$, where $\varphi = \varphi(m)$, be the increasing sequence of all natural numbers not greater than $m$ and coprime with $m$. Then each of the numbers $ar_1, ar_2, \ldots, ar_\varphi$ is also coprime with $m$; let $\rho_1, \rho_2, \ldots, \rho_\varphi$ denote the remainders of these numbers when divided by $m$, so
Each of the numbers $\rho_i$ is coprime with $m$, because according to (1) $\alpha r_i - \varphi_i$ is divisible by $m$, and $\alpha r_i$ is coprime with $m$. Moreover, the numbers $\rho_i$ are all different, because from the equality $\rho_i = \rho_k$ it would follow that $\alpha r_i \equiv \alpha r_k \pmod m$, and thus $r_i \equiv r_k \pmod m$, i.e., $r_i \equiv r_k$, contrary to the definition of the numbers $r_i$. Therefore, the numbers $\rho_1, \rho_2, \ldots, \rho_\varphi$ differ from the numbers $r_1, r_2, \ldots, r_\varphi$ at most in order, and the equality
holds.
Multiplying the congruences (1) side by side, we get
Since each of the numbers $r_i$ is coprime with $m$, the same applies to the product of these numbers, so from (2) and (3) it follows that
which was to be proved.
Note that if $m$ is a prime number, then $\varphi(m) = m - 1$; in this case, Euler's Theorem gives Fermat's Little Theorem. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 865 |
XXXV OM - I - Problem 9
Three events satisfy the conditions:
a) their probabilities are equal,
b) any two of them are independent,
c) they do not occur simultaneously.
Determine the maximum value of the probability of each of these events. | om35_1r_img_4.jpg
Suppose that $ A $, $ B $, $ C $ are these events, each with a probability of $ p $. Given the assumption of independence between any two of the considered events, the probability of each intersection $ A \cap B $, $ B \cap C $, $ C \cap A $ is $ p^2 $. Furthermore, $ A \cap B \cap C \ne 0 $. Therefore,
Since of course $ P (A \cup B \cup C) \geq P (A \cap B) + P (B \cap C) + P (C \cap A) = 3p^2 $, it follows that $ 3p-3p^2 \geq 3p^2 $, which implies that $ p \leq \frac{1}{2} $.
The value $ p = \frac{1}{2} $ is assumed, for example, by the following events. When flipping a coin twice, we take $ A $ to be the event that a tail appears on the first flip, $ B $ - that a head appears on the second flip, $ C $ - that the same result appears on both flips. Clearly, $ P(A) = P(B) = P(C) = \frac{1}{2} $, the events $ A $, $ B $, $ C $ are pairwise independent and cannot occur simultaneously.
Therefore, the maximum value of the considered probability is $ \frac{1}{2} $. | \frac{1}{2} | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 866 |
XXXIV OM - III - Problem 6
Prove that if all dihedral angles of a tetrahedron are acute, then all its faces are acute triangles. | We assign a unit vector perpendicular to each face and directed outward from the tetrahedron. Consider two faces forming an angle $ \alpha $. The vectors assigned to these faces form an angle $ \pi - \alpha $ (Fig. 12). Therefore, for all dihedral angles of the tetrahedron to be acute, it is necessary and sufficient that the angles between each pair of the vectors $ \overrightarrow{a} $, $ \overrightarrow{b} $, $ \overrightarrow{c} $, $ \overrightarrow{d} $ assigned to the respective faces be obtuse, and thus that
om34_3r_img_12.jpg
om34_3r_img_13.jpg
Assume that the above inequalities are satisfied. We will prove that every planar angle of the tetrahedron is acute. Consider the angle $ ABC $. The vector $ \overrightarrow{a} $ assigned to the face $ BCD $ is perpendicular to the line $ BC $. Therefore, the projection $ \overrightarrow{a}_{ABC} $ of the vector $ \overrightarrow{a} $ onto the plane $ ABC $ is also perpendicular to the line $ BC $. Similarly, the vector $ \overrightarrow{c} $ assigned to the face $ ABD $ is perpendicular to the line $ AB $, so its projection $ \overrightarrow{c}_{ABC} $ onto the plane $ ABC $ is perpendicular to $ AB $. It follows that
Therefore, the angle $ ABC $ is acute if and only if $ \overrightarrow{a}_{ABC} \cdot \overrightarrow{c}_{ABC} < 0 $.
om34_3r_img_14.jpg
The vectors $ \overrightarrow{a} $, $ \overrightarrow{b} $, $ \overrightarrow{c} $, $ \overrightarrow{d} $ have length $ 1 $, so (Fig. 14)
Therefore,
since each of the scalar products appearing here is negative. We have thus shown that the angle $ ABC $ is acute. The proof for the remaining planar angles is analogous. | proof | Geometry | proof | Yes | Yes | olympiads | false | 867 |
LIX OM - III - Task 1
In the fields of an $ n \times n $ table, the numbers $ 1, 2, \dots , n^2 $ are written, with the numbers $ 1, 2, \dots , n $ being in the first row (from left to right), the numbers $ n +1, n +2, \dots ,2n $ in the second row, and so on.
$n$ fields of the table have been selected, such that no two lie in the same row or column. Let $ a_i $ be the number found in the selected field that lies in the row numbered $ i $. Prove that | We will first prove that
Let $ b_i $ denote the column number in which the number $ a_i $ is located. Then,
From the conditions of the problem, it follows that the sequence $ (b_1,b_2,\dots ,b_n) $ is a permutation of the sequence $ (1, 2,\dots ,n) $. Therefore,
and in view of formula (2), the proof of relation (1) reduces to a calculation:
Using now equality (1) and the inequality between the arithmetic mean and the harmonic mean, we obtain
which is the thesis of the problem. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 872 |
XVIII OM - III - Problem 3
In a room, there are 100 people, each of whom knows at least 67 others. Prove that there is a quartet of people in the room where every two people know each other. We assume that if person $A$ knows person $B$, then person $B$ also knows person $A$. | In the set $M$ of people in the room, there are at most $32$ people who do not know a certain person $A$ (assuming that person $A$ knows themselves).
If $A$, $B$, and $C$ are any three people in the set $M$, then there is a person $D$ in the set $M$, different from $A$, $B$, and $C$, who knows these three people. Indeed, if all the people who do not know $A$, all the people who do not know $B$, and all the people who do not know $C$ leave the room, then at most $3 \cdot 32 = 96$ people will leave. Each of the remaining people then knows $A$, $B$, and $C$, and since there are at least $100-96 = 4$ of them, at least one of them is different from $A$, $B$, and $C$.
We can thus easily find a quartet of people in the set $M$ who know each other. Specifically, $A$ can be any person in this set, $B$ a person who knows $A$ and is different from $A$, $C$ a person who knows $A$ and $B$ and is different from $A$ and $B$, and finally $D$ a person who knows $A$, $B$, and $C$ and is different from $A$, $B$, and $C$.
Note. In a similar way, we can prove a more general theorem: If there are $n$ people in the room, each of whom knows at least $\left[ \frac{2n}{3} \right] +1$ of the remaining people, then there is a quartet of people in the room, in which every two people know each other.
To prove this, we first note that the number of people who do not know a certain person is at most $n - \left[ \frac{2n}{3} \right] -2$. The number of people who do not know at least one of the three people $A$, $B$, and $C$ is therefore at most $m = 3n - 3 \left[ \frac{2n}{3} \right] - 6$. Since $\left[ \frac{2n}{3} \right] > \frac{2n}{3} -1$, it follows that $3 \left[ \frac{2n}{3} \right] > 2n - 3$, so $3 \left[ \frac{2n}{3} \right] \geq 2n-2$, and thus $m \leq 3n - (2n-2) - 6 = n-4$. The number of people who know $A$, $B$, and $C$ is therefore at least $n-m \geq 4$, so at least one of these people is different from $A$, $B$, and $C$. The rest of the proof remains the same as before. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 874 |
XLII OM - I - Problem 1
On the shore of a lake in the shape of a circle, there are four piers $K$, $L$, $P$, $Q$. A kayak departs from pier $K$ heading towards pier $Q$, and a boat departs from pier $L$ heading towards pier $P$. It is known that if, maintaining their speeds, the kayak headed towards pier $P$ and the boat towards pier $Q$, a collision would occur. Prove that the kayak and the boat will reach their destinations at the same time. | om42_1r_img_1.jpg
Suppose the kayak is moving at a speed of $ v_k $, and the boat is moving at a speed of $ v_l $. Let $ X $ be the point where a "collision" would occur. The lengths of the segments into which point $ X $ divides the chords $ KP $ and $ LQ $ are related by $ |KX| \cdot |PX| = |LX| \cdot |QX| $ (Figure 1).
From the conditions of the problem, it follows that the ratio of speeds $ v_k \colon v_l = |KX| \colon |LX| $. From these two equalities, we obtain the proportion $ v_k \colon v_l = |QX| \colon |PX| $.
The vertical angles $ KXQ $ and $ LXP $ are equal. Therefore, triangles $ KXQ $ and $ LXP $ are similar in the ratio $ v_k \colon v_l $, and thus $ |KQ| \colon |LP| = v_k \colon v_l $, that is,
The left side of the last proportion expresses the time used by the kayak to travel from $ K $ to $ Q $; the right side - the time for the boat to travel from $ L $ to $ P $. Hence the thesis. | proof | Geometry | proof | Yes | Yes | olympiads | false | 875 |
LIX OM - III - Task 6
Let $ S $ be the set of all positive integers that can be represented in the form $ a^2 + 5b^2 $
for some relatively prime integers $ a $ and $ b $. Furthermore, let $ p $ be a prime number that gives a remainder
of 3 when divided by 4. Prove that if some positive multiple of the number $ p $ belongs to the set $ S $, then the number
$ 2p $ also belongs to the set $ S $. | Given the conditions of the problem, the number $a^2 + 5b^2$ is divisible by $p$ for some relatively prime integers $a$ and $b$.
The number $b$ is not divisible by $p$; otherwise, from the divisibility $p \mid b$ and $p \mid a^2 + 5b^2$, we would obtain $p \mid a$, contradicting the relative primality of $a$ and $b$.
Therefore, none of the $p - 1$ numbers: $b, 2b, \dots, (p - 1)b$ is divisible by $p$. Moreover, these numbers give different remainders when divided by $p$. Indeed, if we have $p \mid ib - jb = (i - j)b$ for some indices $1 \leqslant i, j \leqslant p - 1$, then the prime number $p$ is a divisor of the difference $i - j$, implying that $i = j$. Consequently, one of the considered numbers gives a remainder of 1 when divided by $p$. Let us assume that this number is $kb$ (for some value $k \in \{1, 2, \dots, p - 1\}$). Then we have
Let $z$ denote the largest integer not exceeding $p$. Consider the set of remainders when the numbers $0, m, 2m, \dots, zm$ are divided by $p$. Arrange these $z + 1$ remainders in non-decreasing order to obtain the sequence $r_1 \leqslant r_2 \leqslant \dots \leqslant r_{z+1}$. Then each of the $z + 1$ numbers:
is non-negative, and their sum is $p$. It follows that at least one of these numbers is no greater than
Thus, among the numbers $r_2 - r_1, r_3 - r_2, \dots, r_{z+1} - r_z, p - (r_{z+1} - r_1)$, we can find a number not exceeding $z$. This means that for some two different numbers, say $cm$ and $dm$ ($0 \leqslant c, d \leqslant z$), their remainders when divided by $p$ differ by no more than $z$, or differ by at least $p - z$. Let $y = |c - d|$; then one of the numbers $ym, -ym$ gives a remainder $x \leqslant z$ when divided by $p$. Since $p \mid m^2 + 5$, the numbers
give the same remainder when divided by $p$. One of the factors $x - my, x + my$ is divisible by $p$. Therefore, $p \mid x^2 + 5y^2$. Moreover, $0 \leqslant x \leqslant z$ and $0 \leqslant y \leqslant z$, which, combined with the inequality $z < p$, gives
Hence, we conclude that $x^2 + 5y^2$ is one of the numbers $p, 2p, 3p, 4p$, or $5p$.
The equality $x^2 + 5y^2 = p$ would mean that the number $x^2 + 5y^2$ gives a remainder of 3 when divided by 4. This is not possible, since the numbers $x$ and $5y^2$ give remainders of 0 or 1 when divided by 4. Therefore, $x^2 + 5y^2 = p$. This also proves that the equalities $x^2 + 5y^2 = 4p$ and $x^2 + 5y^2 = 5p$ cannot hold. The first of these equalities would mean that the numbers $x$ and $y$ are even (in no other case is $x^2 + 5y^2$ divisible by 4) and
\[
\left(\frac{1}{2}x\right)^2 + 5\left(\frac{1}{2}y\right)^2 = p;
\]
the second equality leads to the conclusion that $5 \mid x$ and
\[
y^2 + 5\left(\frac{1}{2}x\right)^2 = p.
\]
In both cases, we arrive at a contradiction with the previously proven fact that a number of the form $e^2 + 5f^2$ is different from $p$ for any integer values of $e$ and $f$.
Thus, we have established that $x^2 + 5y^2 = 2p$ or $x^2 + 5y^2 = 3p$.
Suppose that $x^2 + 5y^2 = 3p$. Then the numbers $x$ and $y$ are not divisible by 3 (otherwise, we would have $3 \mid x$ and $3 \mid y$, but then $3^2 \mid x^2 + 5y^2 = 3p$; this can only happen if $p = 3$, but then we obtain the following contradiction: $9 = 3p = x^2 + 5y^2 \geqslant 3^2 + 5 \cdot 3^2 = 54$). By changing the signs of the numbers $x$ and $y$ if necessary, we can assume that $x$ and $y$ are integers giving a remainder of 1 when divided by 3. Moreover,
The numbers $x + 5y$ and $x - y$ are divisible by 3; if $x + 5y = 3g$ and $x - y = 3h$, then from the above equality, we conclude that $2p = g^2 + 5h^2$.
Ultimately, we have proven that the number $2p$ can be expressed in the form $s^2 + 5t^2$ for some integers $s$ and $t$. It remains to note that in this case, the numbers $s$ and $t$ are relatively prime; if they had a common prime divisor $q$, we would obtain $q^2 \mid s^2 + 5t^2 = 2p$, leading to a contradiction, since $p$ is an odd prime. Thus, the proof of the relation $2p \in S$ is complete. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 876 |
XXVII OM - II - Problem 3
We consider a hemispherical cap that does not contain any great circle. The distance between points $A$ and $B$ on such a cap is defined as the length of the arc of the great circle of the sphere with endpoints at points $A$ and $B$, which is contained in the cap. Prove that there is no isometry mapping this cap onto a subset of the plane.
Note. A hemispherical cap is any of the two parts into which a plane intersecting a sphere divides its surface. | Let a given spherical cap correspond to the central angle $ \alpha $. From the assumption, we have $ 0 < \alpha < \frac{\pi}{2} $. For any angle $ \beta $ satisfying $ 0 < \beta < \alpha $, consider a regular pyramid $ OABCD $, whose base is a square $ ABCD $ inscribed in the given spherical cap, and whose vertex $ O $ is the center of the sphere containing this cap, such that $ \measuredangle AOC = \beta $ (Fig. 13). Let $ \measuredangle AOD = \gamma $ and let points $ P $ and $ Q $ be the midpoints of segments $ \overline{AD} $ and $ \overline{AC} $, respectively.
We have the following relationships
Therefore
If $ R $ is the radius of the sphere containing the given spherical cap, then the distances between points $ A $ and $ C $ and points $ B $ and $ D $ on the cap are equal to $ R\beta $, and the distances between points $ A $ and $ B $, $ B $ and $ C $, $ C $ and $ D $, and $ D $ and $ A $ on the cap are equal to $ R\gamma $.
If there exists an isometry $ \varphi $ of the given spherical cap onto a subset of the plane, then $ \varphi(A) \varphi (B)\varphi (C) \varphi (D) $ is a quadrilateral on the plane such that $ \varphi(A)\varphi(B) = \varphi(B)\varphi(C) = \varphi(C)\varphi(D) = \varphi(D)\varphi(A) = R \gamma $. This is therefore a rhombus. Its diagonals have equal lengths: $ \varphi(A)\varphi(C) = \varphi(B) \varphi(D) = R\beta $. The considered quadrilateral is thus a square. Therefore, $ \varphi(A)\varphi(C) = \sqrt{2}\varphi(A)\varphi(B) $, i.e.,
From (1) and (2), it follows that $ \sin \frac{\beta}{2} = \sqrt{2} \sin \frac{\beta}{2\sqrt{2}} $ for every angle $ \beta $ satisfying $ 0 < \beta < \alpha $. Substituting in particular the angle $ \frac{\beta}{\sqrt{2}} $ for $ \beta $ here, we get $ \sin \frac{\beta}{2\sqrt{2}}= \sqrt{2} \sin \frac{\beta}{4} $. From the last two equalities, it follows that $ \sin \frac{\beta}{2} =2 \sin \frac{\beta}{4} $. On the other hand, we have $ \sin \frac{\beta}{2} = 2 \sin \frac{\beta}{4} \cos \frac{\beta}{4} $. Comparing the obtained results, we get that $ \cos \frac{\beta}{4} = 1 $. This is impossible, since $ 0 < \frac{\beta}{4} < \frac{\alpha}{4} < \frac{\pi}{8} $.
The obtained contradiction proves that there is no isometry $ \varphi $ of the given spherical cap onto a subset of the plane. | proof | Geometry | proof | Yes | Yes | olympiads | false | 877 |
XXVIII - I - Problem 8
Prove that the set $ \{1, 2, \ldots, 2^{s+1}\} $ can be partitioned into two $ 2s $-element sets $ \{x_1, x_2, \ldots, x_{2^s}\} $, $ \{y_1, y_2, \ldots, y_{2^s}\} $, such that for every natural number $ j \leq s $ the following equality holds | We will apply induction with respect to $s$. For $s = 1$, we have the set $\{1, 2, 3, 4\}$. By splitting it into subsets $\{1, 4\}$ and $\{2, 3\}$, we will have the conditions of the problem satisfied, since $1 + 4 = 2 + 3$.
Next, assume that for some natural number $s$, there exists such a partition of the set $\{1, 2, \ldots, 2^{s+1}\}$ into subsets $\{x_1, x_2, \ldots, x_{2^s}\}$ and $\{y_1, y_2, \ldots, y_{2^s}\}$, that
\[
\sum_{i=1}^{2^s} x_i^r = \sum_{i=1}^{2^s} y_i^r
\]
for $r = 1, 2, \ldots, s + 1$.
We will prove that the partition of the set $\{1, 2, \ldots, 2^{s+2}\}$ into subsets
\[
\{x_1, x_2, \ldots, x_{2^s}, x_1 + 2^{s+1}, x_2 + 2^{s+1}, \ldots, x_{2^s} + 2^{s+1}\}
\]
and
\[
\{y_1, y_2, \ldots, y_{2^s}, y_1 + 2^{s+1}, y_2 + 2^{s+1}, \ldots, y_{2^s} + 2^{s+1}\}
\]
satisfies the conditions of the problem.
For $r = 1, 2, \ldots, s + 1$, by the binomial formula, the sum of the $r$-th powers of the elements of the first subset is equal to
\[
\sum_{i=1}^{2^s} (x_i + 2^{s+1})^r = \sum_{i=1}^{2^s} \sum_{k=0}^r \binom{r}{k} x_i^k (2^{s+1})^{r-k}
\]
Similarly, we calculate the sum of the $r$-th powers of the elements of the second subset:
\[
\sum_{i=1}^{2^s} (y_i + 2^{s+1})^r = \sum_{i=1}^{2^s} \sum_{k=0}^r \binom{r}{k} y_i^k (2^{s+1})^{r-k}
\]
Comparing the obtained results, we see that they differ only in that one formula contains the sum
\[
\sum_{i=1}^{2^s} x_i^k
\]
and the other contains the sum
\[
\sum_{i=1}^{2^s} y_i^k
\]
From formula (1), it follows that these sums are equal. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 879 |
VII OM - I - Task 3
In a square $ABCD$ with area $S$, vertex $A$ is connected to the midpoint of side $BG$, vertex $B$ is connected to the midpoint of side $CD$, vertex $C$ is connected to the midpoint of side $DA$, and vertex $D$ is connected to the midpoint of side $AB$. Calculate the area of the part of the square that contains its center. | The solution to the problem is illustrated in Fig. 1, where $ M $, $ N $, $ P $, $ Q $ denote the midpoints of the sides, and $ O $ the center of the square; through the vertices $ A $, $ B $, $ C $, $ D $ lines parallel to the lines $ DQ $, $ AM $, $ BN $, $ CP $ have been drawn. If the figure is rotated by $ 90^\circ $ around the point $ O $, point $ A $ will fall on point $ B $, point $ Q $ on point $ M $, etc., and the figure will overlap with itself. From this, it is easy to conclude that the figure consists of a grid made up of 9 equal squares. According to the Pythagorean theorem, the area of square $ ABCD $ equals the sum of the areas of the squares constructed on the legs $ AK $ and $ BK $ of the right triangle $ AKB $. Denoting the sought area by the letter $ x $, we therefore have | \frac | Geometry | math-word-problem | Yes | Yes | olympiads | false | 880 |
VI OM - I - Problem 10
Prove that the number $ 53^{53} - 33^{33} $ is divisible by $ 10 $. | It is enough to prove that the last digits of the numbers $53^{53}$ and $33^{33}$ are the same. The last digit of the product $(10k + a)(10m + b)(10n + c) \ldots$, where $k, m, n, \ldots a, b, c, \ldots$ are natural numbers, is equal to the last digit of the product $abc\ldots$. The number $53^{53}$ therefore has the same last digit as the number $3^{53}$. But $3^{53} = 3^{52} \cdot 3 = (81)^{13} \cdot 3$, so the number $3^{53}$ has the same last digit as the number $1^{13} \cdot 3$, i.e., the digit $3$. Similarly, the number $33^{33}$ has the same last digit as the number $3^{33} = 3^{32} \cdot 3 = (3^4)^8 \cdot 3 = (81)^8 \cdot 3$, i.e., the digit $3$, q.e.d.
The above reasoning can be given a concise and clear form using the concepts and simplest properties of congruences (see Problems from Mathematical Olympiads, problem No. 8).
Since $53 \equiv 3 \pmod{10}$, $33 \equiv 3 \pmod{10}$, $3^4 \equiv 1 \pmod{10}$, therefore
\[53^{53} \equiv 3^{53} \equiv (3^4)^{13} \cdot 3 \equiv 3 \pmod{10},\]
\[33^{33} \equiv 3^{33} \equiv (3^4)^8 \cdot 3 \equiv 3 \pmod{10},\]
therefore | proof | Number Theory | proof | Yes | Yes | olympiads | false | 881 |
XIX OM - I - Problem 6
Prove that if the polynomial $ F(x) $ with integer coefficients has a root $ \frac{p}{q} $, where $ p $ and $ q $ are relatively prime integers, then $ p - q $ is a divisor of the number $ F(1) $, and $ p + q $ is a divisor of the number $ F(-1) $. | Let us substitute $ x $ with the variable $ y = qx $. Equation (1) then takes the form
Consider the polynomial
whose coefficients are integers. It has an integer root $ y=p $, hence
where $ Q(y) $ is a polynomial with integer coefficients*).
Substituting $ y = q $ into this equality, we get
The numbers $ q^n $ and $ q - p $ are coprime; if some prime number were a common divisor of $ q^n $ and $ q - p $, it would also be a common divisor of $ q $ and $ q - p $, and thus also a common divisor of $ q $ and $ q-(q-p)=p $, contradicting the assumption.
Therefore, the number $ q - p $ is a divisor of the number $ F(1) $, Q.E.D.
Note. A shorter solution to the problem can be obtained by relying on the following theorem:
If a polynomial $ F(x) $ with integer coefficients has a rational root $ \frac{q}{p} $ ($ p $, $ q $ - integers that are coprime), then the quotient of the polynomial $ F(x) $ divided by $ x - \frac{p}{q} $ is a polynomial with integer coefficients.
We will prove this theorem. Let
where the numbers $ a_0, a_1, \ldots,a_n $ are integers. We need to prove that the numbers $ c_0, c_1, \ldots, c_{n-1} $ are integers.
The coefficients of $ x^n $ on both sides of the equality (*) are equal, so $ c_0 = a_0 $.
Notice that
which implies that $ a_0 $, and thus $ c_0 $, is divisible by $ q $. We will prove the theorem by induction.
When $ n = 1 $, the thesis of the theorem is obvious; the quotient of $ F(x) $ divided by $ x - \frac{p}{q} $ is then equal to $ c_0 $.
Suppose the thesis of the theorem is true for some $ n $ and let
Form the polynomial of degree $ n $
Since the number $ c_0 = a_0 $ is divisible by $ q $, it is a polynomial with integer coefficients. According to the induction hypothesis, the numbers $ c_1, c_2, \ldots, c_n $ are integers, i.e., the thesis of the theorem is true for polynomials of degree $ n + 1 $.
The theorem is proved.
The solution to the previous problem is now immediate. If
and the polynomial $ F(x) $ has integer coefficients, then the polynomial $ Q(x) $ also has integer coefficients and
thus
which implies that the number $ p - q $ is a divisor of the number $ F(1) $. | proof | Algebra | proof | Yes | Yes | olympiads | false | 882 |
XLVII OM - I - Problem 5
In the plane, a triangle $ ABC $ is given, where $ |\measuredangle CAB| = \alpha > 90^{\circ} $, and a segment $ PQ $, whose midpoint is point $ A $. Prove that | Let $ M $ be the midpoint of side $ BC $ of triangle $ ABC $, and $ O $ - the center of the circle $ \omega $ circumscribed around this triangle. Angle $ CAB $ is obtuse by assumption. It follows that points $ A $ and $ O $ lie on opposite sides of line $ BC $; in particular, point $ O $ does not coincide with $ M $.
Let $ ON $ be the radius of circle $ \omega $ passing through point $ M $ (see figure 4). The circle with center $ M $ and radius $ MN $ is tangent to circle $ \omega $ at point $ N $, and point $ A $ lies outside circle $ \omega $ (or on it - when it coincides with $ N $). Therefore, $ |MA| \geq |MN| $.
om47_1r_img_4.jpg
Points $ B $ and $ C $ are symmetric with respect to line $ NO $, which, therefore, bisects angle $ BNC $. And since $ | \measuredangle BNC | = |\measuredangle BAC| = \alpha $ (angles inscribed in the same arc $ BC $), we get the equality
On the extension of side $ CA $, we lay off a segment $ AD $ equal in length to it. Triangle $ BCD $ is formed; points $ A $ and $ M $ are the midpoints of its sides $ CD $ and $ CB $. Therefore, triangles $ BCD $ and $ MCA $ are similar. We obtain the relationship
Point $ A $ is the common midpoint of segments $ PQ $ and $ CD $. Therefore, quadrilateral $ CPDQ $ is a parallelogram (which can degenerate into a segment; note that the reasoning in this part is completely independent of the positions of points $ P $ and $ Q $ relative to the previously considered segments and circles). Thus, $ |CQ|=|PD| $, from which we get
We multiply the inequalities (1) and (2) side by side and obtain the inequality
which was to be proven. | proof | Geometry | proof | Yes | Yes | olympiads | false | 883 |
XXIII OM - III - Problem 4
On a straight line without common points with the sphere $K$, points $A$ and $B$ are given. The orthogonal projection $P$ of the center of the sphere $K$ onto the line $AB$ lies between points $A$ and $B$, and $AP$ and $BP$ are greater than the radius of the sphere. We consider the set $Z$ of triangles $ABC$ whose sides $\overline{AC}$ and $\overline{BC}$ are tangent to the sphere $K$. Prove that $T$ is a triangle with the largest perimeter among the triangles in the set $Z$ if and only if $T$ is a triangle with the largest area among the triangles in the set $Z$. | We will first prove the
Lemma. If $ S $ is a vertex, $ O $ is the center of the base of a right circular cone, and point $ T $ lies in the plane of the base, then the line $ ST $ forms the largest angle with such a generatrix $ SR $ of the cone that $ O \in \overline{TR} $ (Fig. 16).
Proof. For any point $ R $ on the circumference of the base of the cone, by the Law of Cosines, we have
Since the lengths $ SR $ and $ TS $ are constant (do not depend on the position of point $ R $), the angle $ \measuredangle TSR $ will be the largest when its cosine is the smallest, i.e., when $ TR $ is as large as possible. This occurs when $ O \in \overline{TR} $.
We now proceed to solve the problem. Let $ ABC_O $ be the triangle in the family $ Z $ that contains the center $ O $ of the sphere $ K $ (this triangle certainly belongs to the family $ Z $).
Let $ \alpha_0 = \measuredangle BAC_0 $, $ \beta_0 = \measuredangle ABC_0 $, and for any triangle $ ABC $ in the family $ Z $, let $ \alpha = \measuredangle BAC $, $ \beta = \measuredangle ABC $. Since the line $ AC $ is tangent to the sphere $ K $, it is a generatrix of a right circular cone with axis $ AO $. By the lemma, we have $ \alpha_0 \geq \alpha $ and similarly $ \beta_0 \geq \beta $. Drawing the triangles $ ABC_0 $ and $ ABC $ in the same plane (Fig. 17), we observe that $ \triangle ABC \subset \triangle ABC_0 $ and the triangles share a common base. Therefore, the area and perimeter of triangle $ ABC_0 $ are not less than the area and perimeter of triangle $ ABC $, respectively. | proof | Geometry | proof | Yes | Yes | olympiads | false | 885 |
X OM - III - Task 5
In the plane of triangle $ ABC $, a line moves, intersecting sides $ AC $ and $ BC $ at points $ D $ and $ E $ such that $ AD = BE $. Find the geometric locus of the midpoint $ M $ of segment $ DE $. | When $ AC = BC $, the sought geometric locus is the angle bisector of $ C $. It remains to consider only the case where $ AC \ne BC $. We will further assume that $ AC < BC $.
First, note that when $ AC < BC $, the line $ DE $ intersects the line $ AB $ at a point lying on the extension of $ AB $ beyond point $ B $, since point $ E $ lies in the strip between the line $ AB $ and the parallel to $ AB $ passing through point $ D $ (which we assert by invoking Thales' theorem). Hence, we conclude that $ \measuredangle CDE > \measuredangle CAB $ (by the external angle theorem of a triangle), so $ \measuredangle CED < \measuredangle CBA $ (Fig. 32).
We draw through points $ A $ and $ M $ lines parallel to lines $ DE $ and $ AC $ respectively, and obtain parallelogram $ ADMK $, where point $ K $ lies outside triangle $ ABC $. Similarly, we construct parallelogram $ BEMN $. Vertex $ N $ lies inside triangle $ ABC $. Therefore, segment $ KN $ intersects segment $ AB $ at some point $ S $. Triangles $ AKS $ and $ BNS $ are symmetric with respect to point $ S $, since, due to the parallelism of lines $ AK $ and $ NB $, they are similar with respect to point $ S $, and moreover, $ AK = DM = ME = NB $. It follows that point $ S $ is the midpoint of both segments $ AB $ and $ KN $. Triangle $ KMN $ is isosceles, since $ KM = AD = BE = NM $, so line $ MS $ is the angle bisector of $ KMN $. Since the sides of angle $ KMN $ are correspondingly parallel to the sides of angle $ ACB $, line $ MS $ is parallel to the angle bisector of $ ACB $.
We have proved that the point $ M $, whose geometric locus we are seeking, lies on the line passing through the midpoint $ S $ of segment $ AB $ and parallel to the angle bisector of $ ACB $, and specifically within the segment $ SO $ of this line lying inside triangle $ ABC $. Conversely, any interior point of segment $ SO $ of this line belongs to the sought geometric locus. We easily verify this by reversing the previous reasoning. Let $ M $ be any interior point of segment $ SO $. We draw through point $ M $ lines parallel to $ CA $ and $ CB $, and through point $ S $ a perpendicular to $ SO $, and obtain isosceles triangle $ KMN $. Triangles $ ASK $ and $ BSN $ are symmetric with respect to point $ S $, so $ AK \parallel NB $ and $ AK = NB $. Drawing through point $ M $ a line parallel to $ AK $ and $ NB $, we obtain parallelograms $ ADMK $ and $ BEMN $, in which $ AD = KM = NM = BE $ and $ DM = AK = NB = ME $. Indeed, point $ M $ belongs to the geometric locus.
If we choose point $ M $ at point $ O $, then reasoning as above, we will show that point $ O $ is the midpoint of such a segment $ CC $ that $ C $. We can consider that points $ S $ and $ O $ also belong to our geometric locus.
The obtained result can be generalized. If instead of triangle $ ABC $ we consider the entire lines $ AC $ and $ BC $ intersected by a moving line at points $ D $ and $ E $, with $ AD = BE $, then as the geometric locus of the midpoint $ M $ of segment $ DE $, we obtain the entire line $ SO $. | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 886 |
LIV OM - II - Task 4
Prove that for every prime number $ p > 3 $ there exist integers $ x $, $ y $, $ k $ satisfying the conditions: $ 0 < 2k < p $ and | Let $ A $ be the set of numbers of the form $ x^2 $, where $ 0 \leq x < p/2 $, and $ B $ the set of numbers of the form $ 3 - y^2 $, where $ 0 \leq y < p/2 $; the numbers $ x $, $ y $ are integers. We will show that the numbers in the set $ A $ give different remainders when divided by $ p $.
Suppose that the numbers $ x_1^2 $, $ x_2^2 $ from the set $ A $ give the same remainder when divided by $ p $. Then
from which we have the divisibility
Since $ 0 \leq x_1 $, $ x_2 < p/2 $, we have $ -p/2 < x_1 - x_2 < p/2 $ and $ 0 \leq x_1 + x_2 < p $. Therefore, and from the above divisibility, we conclude that $ x_1 = x_2 $.
Thus, the numbers in the set $ A $ give different remainders when divided by $ p $. Similarly, we prove that the numbers in the set $ B $ give different remainders when divided by $ p $.
Since the sets $ A $ and $ B $ together have $ p + 1 $ elements, there exist two numbers - one from the set $ A $, the other from the set $ B $ - that give the same remainder when divided by $ p $. Let $ x^2 $ and $ 3 - y^2 $ be these numbers. Then the number $ x^2 + y^2 - 3 $ is divisible by $ p $, so we have
for some integer $ k $. Since $ p > 3 $, $ k $ is a positive number and
The conditions given in the problem statement are thus satisfied. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 889 |
XLVIII OM - III - Problem 2
Find all triples of real numbers $ x $, $ y $, $ z $ satisfying the system of equations | Suppose that the numbers $ x $, $ y $, $ z $ satisfy the given system; from the second equation it follows that
Further, we have
which, by the first equation of the system, gives:
There is also the inequality
that is,
By property (1), we can multiply inequalities (3) and (4) side by side:
According to the second equation of the system, inequality (5) must be an equality. This means that either both multiplied inequalities (3) and (4) are equalities, or the expressions on both sides of (4) are equal to zero. In the first case, the equality sign in relations (2) and (3) leads to the conclusion that $ x = y = z = \pm 1/3 $. In the second case, the products $ zy $, $ yz $, $ zx $ are equal to zero, which means that two of the numbers $ z $, $ y $, $ z $ are equal to zero; the third must be equal to $ \pm 1/\sqrt{3} $, in accordance with the first equation of the system. This gives eight triples $ (x, y, z) $ that satisfy the given system of equations:
where $ \varepsilon = \pm 1 $. | (x,y,z)=(\1/3,\1/3,\1/3)(x,y,z)=(\1/\sqrt{3},0,0),(0,\1/\sqrt{3},0),(0,0,\1/\sqrt{3}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 890 |
L OM - I - Task 3
In an isosceles triangle $ ABC $, angle $ BAC $ is a right angle. Point $ D $ lies on side $ BC $, such that $ BD = 2 \cdot CD $. Point $ E $ is the orthogonal projection of point $ B $ onto line $ AD $. Determine the measure of angle $ CED $. | Let's complete the triangle $ABC$ to a square $ABFC$. Assume that line $AD$ intersects side $CF$ at point $P$, and line $BE$ intersects side $AC$ at point $Q$. Since
$ CP= \frac{1}{2} CF $. Using the perpendicularity of lines $AP$ and $BQ$ and the above equality, we get $ CQ= \frac{1}{2} AC $, and consequently $ CP=CQ $. Points $C$, $Q$, $E$, $P$ lie on the same circle, from which $ \measuredangle CED =\measuredangle CEP =\measuredangle CQP = 45^\circ $. | 45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 891 |
XXI OM - III - Task 4
In the plane, there are $ n $ rectangles with sides respectively parallel to two given perpendicular lines. Prove that if any two of these rectangles have at least one point in common, then there exists a point belonging to all the rectangles. | Let us choose a simple set of axes for the coordinate system (Fig. 14). Then the vertices of the $i$-th rectangle $P_i$ will have coordinates $(x_i, y_i)$, $(x_i, y)$, $(x, y_i)$, $(x, y)$, where $x_i < x$, $y_i < y$. The point $(x, y)$ will therefore belong to the rectangle $P_i$ if and only if $x_i \leq x \leq x$ and $y_i \leq y \leq y$.
By assumption, any two rectangles $P_i$ and $P_j$ have a common point. Therefore, there exist numbers $x$, $y$ such that
and
Thus,
Let $a$ be the largest of the numbers $x_i$, and $b$ be the largest of the numbers $y_i$. Then, by definition,
The inequalities (2) and (3) prove that the point $(a, b)$ belongs to each of the rectangles $P_j$.
Note 1. In the given solution, we did not significantly use the fact that the number of given rectangles is finite. If a certain infinite family of rectangles with similar properties were given, the solution would proceed similarly, except that the numbers $a$ and $b$ would be defined as the least upper bounds of the sets of numbers $x_j$ and $y_j$, respectively.
Note 2. A similar theorem is true if, instead of rectangles, we consider any convex figures. Specifically, the following holds:
Helly's Theorem. If on a plane there is a non-empty family of convex sets such that any three sets of this family have a common point, then there exists a point belonging to all sets of this family. | proof | Geometry | proof | Yes | Yes | olympiads | false | 895 |
XXXVIII OM - I - Problem 4
For a given natural number $ n $, we consider the family $ W $ of all sets $ A $ of pairs of natural numbers not greater than $ n $, such that if $ (a, b) \in A $, then $ (x, y) \in A $ for $ 1 \leq x \leq a $, $ 1 \leq y \leq b $.
Prove that the family $ W $ consists of $ \binom{2n}{n} $ different sets. | For any pair of natural numbers $ (x,y) $, we denote by $ Q_{xy} $ the unit square with vertices $ (x-1,y-1) $, $ (x,y-1) $, $ (x,y) $, $ (x-1,y) $, and by $ R_{xy} $ - the rectangle with vertices $ (0,0) $, $ (x,0) $, $ (x,y) $, $ (0,y) $. If $ A $ is any set of pairs of natural numbers, then the symbol $ Z(A) $ will denote the following subset of the plane:
\[
Z(A) = \bigcup_{(x,y) \in A} Q_{xy}
\]
We assume the following:
\[
Z(\emptyset) = \emptyset
\]
According to the problem statement, the set $ A $ belongs to the family $ W $ if and only if the set $ Z = Z(A) $ satisfies the conditions:
1. $ Z $ is the union of a finite number of squares $ Q_{xy} $ ($ x,y\in \mathbb{N} $),
2. $ Z \subset R_{nn} $,
3. $ Q_{ab} \subset Z \Longrightarrow R_{ab} \subset Z $.
In particular, the empty set (consisting of $ 0 $ unit squares) satisfies these conditions.
Assume for a moment that the set $ Z $ is not empty. Implication (3) then expresses exactly that $ Z $ is the union of a certain (finite, positive) number of rectangles of the form $ R_{ab} $ - that is, rectangles with two sides lying on the positive coordinate axes. A finite union of such rectangles (contained in $ R_{nn} $) is a polygon bounded by a closed broken line $ ON_0M_1N_1\ldots M_kN_kO $, where: $ k\in \mathbb{N} $; all vertices have integer coordinates; $ O $ is the origin of the coordinate system; $ N_0 $ is a point on the segment $ OK $, where $ K = (0, n) $; $ N_k $ is a point on the segment $ OL $, where $ L = (n, 0) $; vectors $ \overrightarrow{N_{i-1}M_i} $ are parallel to the $ Ox $ axis and directed in the direction of this axis; vectors $ \overrightarrow{M_iN_i} $ are parallel to the $ Oy $ axis and directed opposite to the direction of this axis ($ i = 1, \ldots , k $) (see figure 3).
om38_1r_img_3.jpg
Conversely, any polygon $ Z $ bounded by such a broken line satisfies conditions (1), (2), (3), and thus corresponds to some non-empty set $ A\in W $. It remains to count these polygons.
Let $ ON_0M_1N_1\ldots M_kN_kO $ be a polygon of the described form. Each of the vectors $ \overrightarrow{N_{i-1}M_i} $ is the sum of a certain number of unit vectors $ [1,0] $, and each of the vectors $ \overrightarrow{M_iN_i} $ is the sum of a certain number of unit vectors $ [0, -1] $. Consider the broken line $ KN_0M_1N_1\ldots M_kN_kL $ (it may be $ K = N_0 $ or $ N_k = L $), oriented from $ K $ to $ L $. It consists of $ 2n $ unit vectors: each of the vectors $ [1, 0] $ and $ [0, -1] $ must appear $ n $ times (going from $ K $ to $ L $ we must make $ n $ steps to the right and $ n $ steps down). The polygon $ Z $ is uniquely determined by a path of this type running from $ K $ to $ L $.
Since we assumed (temporarily) that the set $ Z $ is not empty, we must exclude the path from $ K $ to $ L $ composed sequentially of the vector $ [0, -1] $ repeated $ n $ times, and then the vector $ [1,0] $ repeated $ n $ times (i.e., the broken line $ KOL $). It is natural to now allow this path as well, assuming that it corresponds precisely to the empty set $ Z $ (which must also be included in the considerations).
So how many admissible paths are there from $ K $ to $ L $? We have $ 2n $ unit vectors of two types. If in the sequence $ (1, 2, \ldots, 2n) $ we fix $ n $ positions where we place the vector $ [1, 0] $, then on the remaining $ n $ positions the vector $ [0, -1] $ must appear. There are as many such paths as there are $ n $-element subsets of a $ 2n $-element set, i.e., $ \binom{2n}{n} $. There are as many polygons $ Z $ satisfying conditions (1), (2), (3), and thus as many sets $ A $ in the family $ W $. | \binom{2n}{n} | Combinatorics | proof | Yes | Yes | olympiads | false | 897 |
XXIV OM - I - Problem 11
Prove that if the center of the sphere circumscribed around a tetrahedron coincides with the center of the sphere inscribed in this tetrahedron, then the faces of this tetrahedron are congruent. | Let point $O$ be the center of the inscribed and circumscribed spheres of the tetrahedron $ABCD$. Denote by $r$ and $R$ the radii of these spheres, respectively. If $O$ is the point of tangency of the inscribed sphere with one of the faces, and $P$ is one of the vertices of this face, then applying the Pythagorean theorem to triangle $OO'$ (where $O'$ is the point of tangency), we get $O$. Therefore, the distance from point $O$ to each vertex of this face is the same, i.e., $O$ is the center of the circle circumscribed around this face. It also follows from this that the radii of the circles circumscribed around each face of the tetrahedron have the same length $\sqrt{R^2 - r^2}$.
Point $O$ lies inside the face of the tetrahedron, as it is the point of tangency of the inscribed sphere in the tetrahedron. Since $O$ is also the center of the circle circumscribed around this face, it follows that all angles of this face are acute. These are inscribed angles subtended by arcs smaller than a semicircle (Fig. 12).
Applying the Law of Sines to triangles $ABC$ and $CBD$ (Fig. 13), we get $\sin \measuredangle BAC = \frac{BC}{2 \sqrt{R^2 - r^2}}$ and $\sin \measuredangle BDC = \frac{BC}{2 \sqrt{R^2 - r^2}}$. Therefore, $\sin \measuredangle BAC = \sin \measuredangle BDC$, and since both these angles are acute, $\measuredangle BAC = \measuredangle BDC$. Similarly, we prove that any two angles of the faces of the tetrahedron opposite the same edge are equal, i.e., $\measuredangle ABC = \measuredangle ADC$, $\measuredangle ACB = \measuredangle ADB$, $\measuredangle ABD = \measuredangle ACD$, $\measuredangle BAD = \measuredangle BCD$, $\measuredangle CAD = \measuredangle CBD$. Denoting the common measures of the angles in the last six equalities by $\alpha, \beta, \gamma, \delta, \varepsilon, \mu$ respectively, we have
since the sum of the angles of any face is $\pi$.
Adding the equations (1) and (2) side by side, we get
Subtracting the equations (3) side by side, we obtain $\beta = \varepsilon$. Since $\beta = \measuredangle ABC$ and $\varepsilon = \measuredangle BCD$, it follows that triangles $ABC$ and $DCB$ are congruent (Fig. 13).
Similarly, we prove that any two faces of the tetrahedron $ABCD$ are congruent.
Note. If the center of the circle circumscribed around a triangle is the center of the circle inscribed in this triangle, then the triangle is equilateral. An analogous theorem in space does not hold: There are tetrahedra that are not regular, for which the center of the inscribed sphere is the center of the circumscribed sphere.
Consider, for example, the tetrahedron $ABCD$ where $A = (1, a, 0)$, $B = (-1, a, 0)$, $C = (0, -a, 1)$, $D = (0, -a, -1)$, and $a$ is a positive number different from $\frac{\sqrt{2}}{2}$. Then $AB = 2$ and $AC = \sqrt{2 + 4a^2} \ne 2$. Therefore, the tetrahedron $ABCD$ is not regular. Each of the following isometries $f_1(x, y, z) = (x, y, -z)$, $f_2(x, y, z) = (-x, y, z)$, $f_3(x, y, z) = (z, -y, x)$ maps the set $\{A, B, C, D\}$ onto itself. Therefore, under each of these isometries, the center $P = (r, s, t)$ of the sphere circumscribed around the tetrahedron $ABCD$ maps onto itself. From $f_1(P) = P$ it follows that $t = 0$, from $f_2(P) = P$ that $r = 0$, and from $f_3(P) = P$ that $s = 0$. Therefore, $P = (0, 0, 0)$. Similarly, we conclude that $P$ is the center of the sphere inscribed in the tetrahedron $ABCD$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 900 |
VIII OM - I - Problem 7
Construct triangle $ ABC $ given: side $ AB = c $, circumradius $ R $, and angle $ \delta $ between the angle bisector of $ C $ and the altitude from vertex $ C $. | Let $D$ (Fig. 4) denote the foot of the altitude of triangle $ABC$ drawn from vertex $C$, $E$ - the point of intersection of the angle bisector of angle $C$ with side $AB$, $M$ - the point of intersection of line $CE$ with the circumcircle of triangle $ABC$, and $F$ - the foot of the perpendicular dropped from $M$ to line $AB$. Since $\measuredangle ACM = \measuredangle MCB$, point $M$ is the midpoint of arc $AMB$, and since $CD \parallel FM$, $\measuredangle FME = \measuredangle DCE = \measuredangle \delta$.
From this, we derive the following construction. We draw a circle with a given radius $R$ and draw a chord $AB = c$ in it, which is possible when $c \leq 2R$. Let $M$ be the midpoint of one of the arcs of the circle determined by $AB$. From point $M$, we drop a perpendicular $MF$ to $AB$ and construct angle $FME$ equal to the given angle $\delta$ on the ray $MF$. If the side $ME$ of this angle intersects the drawn circle at point $C$ lying on the opposite side of line $AB$ from $M$, then triangle $ABC$ satisfies the conditions of the problem. Such a point $C$ exists if and only if angle $\delta$ is less than angle $FMA$. Since we can choose point $M$ as the midpoint of either of the two arcs of the circle with chord $AB$ and since we can construct angle $FME$ on either side of ray $MF$, the problem can have 4 solutions symmetric in pairs relative to axis $MF$ or 2 solutions symmetric relative to $MF$, or it may have no solutions at all.
This result can be formulated more precisely. Notice that $\measuredangle FMA = \frac{1}{2} \measuredangle BMA$ and that $\sin \measuredangle BMA = \frac{c}{2R}$. Let $\gamma$ denote such an acute angle that $\sin \gamma = \frac{c}{2R}$, then $\measuredangle BMA = \gamma$ or $\measuredangle BMA = 180^\circ - \gamma$, so $\measuredangle FMA = \frac{\gamma}{2}$ or $\measuredangle FMA = 90^\circ - \frac{\gamma}{2}$. One of three cases may occur:
a) $\delta < \frac{\gamma}{2}$, in which case the problem has 4 solutions symmetric in pairs
b) $\frac{\gamma}{2} \leq \delta < 90^\circ - \frac{\gamma}{2}$, the problem has 2 symmetric solutions
c) $\delta \geq 90^\circ - \frac{\gamma}{2}$, the problem has no solutions.
Note. Let the angles of the triangle be denoted in the usual way by $\alpha$, $\beta$, $\gamma$ and assume that $\alpha > \beta$. Then $\delta = \measuredangle DCE = \measuredangle ACE - \measuredangle ACD = \frac{1}{2} \gamma - (90^\circ - \alpha) = \frac{1}{2} (2 \alpha + \gamma - 90^\circ) = \frac{1}{2} (\alpha - \beta)$. The problem above reduces to solving the system of 2 equations: $\sin (\alpha + \beta) = \frac{c}{2R}$, $\alpha - \beta = 2 \delta$, with 2 unknowns $\alpha$ and $\beta$. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 901 |
II OM - II - Task 6
Given points $ A $ and $ B $ and a circle $ k $. Construct a circle passing through points $ A $ and $ B $ and intersecting circle $ k $ in a common chord of given length $ d $. | Analyze. The task boils down to determining the center $X$ of the common chord of the given circle and the circle to be found; for, having the point $X$, we can draw in the given circle $k$ a chord $CD$ with the center $X$ and draw a circle through the points $A$, $B$, $C$, $D$.
The geometric locus of the centers $X$ of the chords of the circle $k$ having a given length $d$ is a circle $k$ concentric with the circle $k$ and having a radius $\sqrt{r^2 - \frac{d^2}{4}}$, where $r$ denotes the radius of the circle $k$. It should be noted, however, that $d \leq 2r$.
1° If the given points $A$ and $B$ are equidistant from the center $O$ of the circle $k$ (Fig. 57), then the circle sought intersects the circle $k$ at points symmetric with respect to the perpendicular bisector of the segment $AB$, and thus the point $X$ will be found at the intersection of this perpendicular bisector with the circle $k$.
2° If $A$ and $B$ are not equidistant from the point $O$ (Fig. 58), then all common chords of the circle $k$ with circles passing through the points $A$ and $B$ lie on lines that intersect the line $AB$ at the same point $M$ (see problem no. 22 on page 105). The point $M$ can be found by drawing through the points $A$ and $B$ an auxiliary circle $l$ intersecting the circle $k$. The point $X$ will be found at the intersection of the circle $k$ with the circle having the diameter $OM$, which is the geometric locus of the mentioned chords.
Construction. We draw the circle $k$ in the usual way.
In case 1°, we draw the perpendicular bisector of the segment $AB$. We draw through the point of intersection $X$ of this perpendicular bisector with the circle $k$ a chord $CD$ of the circle $k$ perpendicular to the line $O$. We then draw a circle through the points $A$, $B$, $C$; this circle will also pass through the point $D$, as the point $D$ is symmetric to the point $C$ with respect to the line $OX$. The circle drawn has a common chord $CD$ of length $d$ with the circle $k$, and is thus the circle sought in the problem.
In case 2°, we draw through the points $A$ and $B$ an arbitrary circle $l$ intersecting the circle $k$ at points $P$ and $Q$. The line $PQ$ intersects the line $AB$ at the point $M$. We draw a circle with the diameter $OM$.
If $X$ is the point of intersection of this circle with the circle $k$, then the line $MX$ intersects the circle $k$ along a chord $CD$ of length $d$. We now draw a circle through the points $A$, $B$, $C$; this circle will also pass through the point $D$, since $MA \cdot MB = MP \cdot MQ$, $MP \cdot MQ = MC \cdot MD$, so $MA \cdot MB = MC \cdot MD$. This will be the circle sought in the problem.
Discussion. We will examine the feasibility of the construction and the number of solutions to the problem depending on the choice of data. We have already noted that the problem can be solved only if the given length $d$ satisfies the condition $d \leq 2r$, where $r$ is the radius of the given circle $k$.
Assuming that this inequality is satisfied, we will consider all cases that may occur.
I. The center $O$ of the circle $k$ lies on the perpendicular bisector of the segment $AB$.
The perpendicular bisector of the segment $AB$ intersects the circle $k$ at two points $X_1$ and $X_2$, so there are two chords $C_1D_1$ and $C_2D_2$ of the circle $k$ having length $d$ and the direction of the line $AB$.
If the line $AB$ does not coincide with any of the lines $C_1D_1$ and $C_2D_2$, i.e., if the distance of the line $AB$ from the point $O$ does not equal $\sqrt{r^2 - \frac{d^2}{4}}$, the problem has two solutions.
If the line $AB$ coincides, for example, with the line $C_1D_1$, but the points $A$ and $B$ do not lie on the circle $k$, i.e., do not coincide with the points $C_1$ and $D_1$, the problem has one solution, namely the circle passing through the points $A$, $B$, $C_2$, $D_2$.
If finally the points $A$ and $B$ coincide with the points $C_1$ and $D_1$, the solution to the problem is any circle passing through the points $A$ and $B$, except for the circle $k$ itself.
In this case, the circle $k$ reduces to the point $O$, the point $X$ also coincides with the point $O$, and in the circle $k$ there is one chord $CD$ of length $d$ and the direction of the line $AB$.
If the line $AB$ does not pass through the point $D$, the problem has one solution.
If the line $AB$ passes through the point $O$, but the points $A$ and $B$ do not lie on the circle $k$, the problem has no solution.
If finally the segment $AB$ is a diameter of the circle $k$, the solution to the problem is any circle passing through the points $A$ and $B$, except for the circle $k$.
II. The center of the circle $k$ does not lie on the perpendicular bisector of the segment $AB$.
If the line $AB$ lies outside the circle $k$, i.e., if its distance from the point $O$ is greater than $\sqrt{r^2 - \frac{d^2}{4}}$, then the point $M$ also lies outside the circle $k$. The circle with the diameter $OM$ intersects the circle $k$ at two points $X_1$ and $X_2$. The lines $MX_1$ and $MX_2$ determine in the circle $k$ chords $C_1D_1$ and $C_2D_2$ of length $d$. The problem has two solutions: the circle passing through the points $A$, $B$, $C_1$, $D_1$ and the circle passing through the points $A$, $B$, $C_2$, $D_2$.
If the line $AB$ is tangent to the circle $k$ at the point $T$, and the point $M$ is different from the point $T$, then the circle with the diameter $OM$ intersects the circle $k$ at two points $T$ and $X$. In this case, there is only one solution: the circle passing through the points $A$, $B$ and the points $C$, $D$ where the line $MX$ intersects the circle $k$.
If the line $AB$ is tangent to the circle $k$, and the point $M$ coincides with the point of tangency $T$, then there is no solution. This case occurs when the point $T$ lies inside the segment $AB$, and the geometric mean of the segments $AT$ and $BT$ equals $\frac{1}{2} d$.
If finally the line $AB$ intersects the circle $k$, the problem has two solutions, one solution, or no solution, depending on whether the point $M$ lies outside the circle $k$, on this circle, or inside the circle $k$.
In this case, the point $X$ coincides with the point $O$.
If the line $AB$ does not pass through the point $O$, then the line $MO$ intersects the circle $k$ at two points $C$ and $D$. The problem then has one solution: the circle passing through the points $A$, $B$, $C$, $D$.
If the line $AB$ passes through the point $O$, and the point $M$ is different from the point $O$, then the line $MO$ intersects the circle $k$ at the points $C$ and $D$ on the line $AB$ and the problem has no solution.
If finally the point $M$ on the line $AB$ coincides with the point $O$, the solution to the problem is any circle passing through the points $A$ and $B$. This case occurs when the point $O$ lies inside the segment $AB$, and the radius $r$ of the circle $k$ is the geometric mean of the segments $OA$ and $OB$.
Note. In the above problem, the number $d$, as the length of a segment, is a positive number. We can, however, consider the "limiting" case of this problem by taking $d = 0$, which gives the problem:
To draw a circle through two given points $A$ and $B$ that is tangent to a given circle $k$.
The method of solution remains the same - with the difference that in this case the circle $k$ is identical to the circle $k$.
The discussion can easily be derived from the discussion conducted above for the case $d < 2r$.
We discussed this problem in problem no. 22 on page 112. | notfound | Geometry | math-word-problem | Yes | Yes | olympiads | false | 902 |
XXX OM - I - Task 8
Among the cones inscribed in the sphere $ B $, the cone $ S_1 $ was chosen such that the sphere $ K_1 $ inscribed in the cone $ S_1 $ has the maximum volume. Then, a cone $ S_2 $ of maximum volume was inscribed in the sphere $ B $, and a sphere $ K_2 $ was inscribed in the cone $ S_2 $. Determine which is greater: the sum of the volumes of $ S_1 $ and $ K_1 $, or the sum of the volumes of $ S_2 $ and $ K_2 $. | Let $ S $ be any cone inscribed in a given sphere $ B $, and $ K $ - the sphere inscribed in the cone $ S $. The axial section of the cone $ S $ is an isosceles triangle $ ACD $, where $ A $ is the vertex of the cone (Fig. 7).
om30_1r_img_7.jpg
Let $ O $ be the center of the sphere $ B $, and $ Q $ - the center of the sphere $ K $. Then $ O $ is the center of the circle circumscribed around the triangle $ ACD $, and $ Q $ - the center of the circle inscribed in this triangle. Let $ P $ be the midpoint of the segment $ \overline{CD} $, and $ M $ and $ N $ respectively - the orthogonal projections of the points $ Q $ and $ O $ onto the line $ AD $.
Let $ \alpha = \measuredangle PAD $, $ R = AO $. We have the following relationships:
Therefore, the volume $ V_s $ of the cone $ S $ is equal to
or
where we have taken $ a = \sin \alpha $.
We then have $ MQ = QP $. Thus $ AP \sin \alpha = QP \sin \alpha + AQ \sin \alpha = MQ \sin \alpha + MQ $ and therefore
The volume $ V_K $ of the sphere $ K $ is then equal to
Therefore, the sum $ V = V(a) $ of the volumes of the cone $ S $ and the sphere $ K $ is, by (1) and (2):
From the conditions of the problem, it follows that $ 0 < \alpha < \frac{\pi}{2} $. Therefore, the number $ a = \sin \alpha $ belongs to the interval $ (0,1) $.
In this interval, the function $ f(a) = a (1 - a^2) $ attains its maximum value at $ a = \frac{\sqrt{3}}{3} $, since this is the only zero of the derivative of $ f $ in this interval, and $ f(0) = f(1) = 0 $ and $ f(a) > 0 $ for $ 0 < a < 1 $. Therefore, from formula (1), it follows that the cone $ S $ has the maximum volume at $ a = \sin \alpha = \frac{\sqrt{3}}{3} $.
Similarly, we observe that the function $ g(a) = a(1 - a) $ in the interval $ (0, 1) $ attains its maximum value at $ a = \frac{1}{2} $. Therefore, from formula (2), it follows that the sphere $ K $ has the maximum volume at $ a = \sin \alpha = \frac{1}{2} $, i.e., at $ \alpha = \frac{\pi}{6} $.
Next, we calculate the values of the function $ V(a) $ defined by formula (3) at the points $ a = \frac{\sqrt{3}}{3} $ and $ a = \frac{1}{2} $. We obtain:
It is easy to verify that $ \frac{4}{9} (2\sqrt{3} - 3) > \frac{13}{64} $, and thus $ V \left( \frac{\sqrt{3}}{3} \right) > V \left( \frac{1}{2} \right) $. Therefore, the sum of the volumes of $ S_2 $ and $ K_2 $ is greater than the sum of the volumes of $ S_1 $ and $ K_1 $. | V(\frac{\sqrt{3}}{3})>V(\frac{1}{2}) | Geometry | math-word-problem | Yes | Yes | olympiads | false | 904 |
XXXI - I - Task 1
Determine for which values of the parameter $ a $ a rhombus with side length $ a $ is a cross-section of a cube with edge length 2 by a plane passing through the center of the cube. | Suppose the rhombus $KLMN$ is a cross-section of the cube $ABCDA_1B_1C_1D_1$ by a plane passing through the center of the cube $O$. Since $O$ is also the center of symmetry of the cross-section, the opposite vertices of the rhombus lie on opposite edges of the cube. Let, for example, as in the figure, $K \in \overline{AB}$, $L \in \overline{A_1B_1}$, $M \in \overline{C_1D_1}$, $N \in \overline{CD}$. Since $O$ is the center of symmetry of the cube, $AK = MC_1$, $A_1L = NC$. Suppose that $K$ is not the midpoint of the edge $\overline{AB}$. Map the adjacent faces $ABB_1A_1$ and $A_1B_1C_1D_1$ onto the net of the cube.
om31_1r_img_6.jpgom31_1r_img_7.jpg
The right trapezoids $AKLA_1$ and $LB_1C_1M$ are congruent, as they are the sum of congruent rectangles and congruent right triangles. It follows that $A_1L = B_1L$, i.e., $L$ is the midpoint of the segment $\overline{A_1B_1}$, and therefore $N$ is the midpoint of $\overline{CD}$. In this way, we have shown that certain two opposite vertices of the rhombus are the midpoints of the edges of the cube on which they lie. The side of the rhombus is thus a segment connecting the midpoint of an edge with some point on the opposite side of the face, which is a square with side length $2$. The length of such a segment is no less than $2$ and no greater than $\sqrt{5}$, and any value in the interval $[2, \sqrt{5}]$ can be achieved by choosing the point $K$ appropriately on the side $\overline{AB}$. | [2,\sqrt{5}] | Geometry | math-word-problem | Yes | Yes | olympiads | false | 905 |
XIV OM - II - Task 6
From point $ S $ in space, $ 3 $ rays emerge: $ SA $, $ SB $, and $ SC $, none of which is perpendicular to the other two. Through each of these rays, a plane is drawn perpendicular to the plane containing the other two rays. Prove that the planes drawn intersect along a single line $ d $. | In the case where the half-lines lie in the same plane, the theorem is obvious; the line $d$ is then perpendicular to the given plane. Let us assume, therefore, that the half-lines $SA$, $SB$, $SC$ form a trihedral angle. According to the assumption, at most one of the dihedral angles of this trihedron is a right angle; let us assume, for example, that $\measuredangle ASB \ne 90^\circ$ and $\measuredangle ASC \ne 90^\circ$.
Choose an arbitrary point $A_1$ on the edge $SA$ different from $S$ and draw through $A_1$ a plane $\pi$ perpendicular to the line $SA_1$; it intersects the lines $SB$ and $SC$ at some points $B_1$ and $C_1$ (Fig. 25).
Let $\alpha$, $\beta$, $\gamma$ denote the planes drawn through the edges $SA$, $SB$, $SC$ of the trihedron $SABC$ perpendicular to the faces $BSC$, $CSA$, $ASB$, and let $m$, $n$, $p$ be the lines of intersection of the planes $\alpha$, $\beta$, $\gamma$ with the plane $\pi$.
Since $\alpha \bot \pi$ and $\alpha \bot \textrm{plane } BSC$, then $\alpha \bot B_1C_1$, so the line $m$ intersects the line $B_1C_1$ at some point $M$ and is perpendicular to it, $A_1M \bot B_1C_1$. Since $\beta \bot \textrm{plane } ASC$ and $\pi \bot \textrm{plane } ASC$, then $\pi \bot \textrm{plane } ASC$, so $\pi \bot A_1C_1$, and since $n$ and $A_1C_1$ lie in the plane $\pi$, they intersect at some point $N$ and $B_1N \bot A_1C_1$. Similarly, we conclude that the line $p$ intersects the line $A_1B_1$ at some point $P$ and is perpendicular to it, i.e., $C_1P \bot A_1B_1$.
The lines $A_1M$, $B_1N$, and $C_1P$ thus contain the altitudes of the triangle $A_1B_1C_1$, so they intersect at a single point $H$. We have proved that the planes $\alpha$, $\beta$, $\gamma$ have a common line $d = SH$. | proof | Geometry | proof | Yes | Yes | olympiads | false | 906 |
VIII OM - I - Zadanie 5
Jaki warunek powinna spełniać liczbą $ q $, aby istniał trójkąt, którego boki tworzą postęp geometryczny o ilorazie $ q $?
|
Weźmy pod uwagę trzy odcinki, których długości tworzą postęp geometryczny o ilorazie $ q $. Ponieważ stosunek dwóch odcinków jest liczbą niezależną od jednostki, jaką te odcinki mierzymy, więc jeśli za jednostkę obierzemy pierwszy z rozważanych trzech odcinków, to długości tych odcinków wyrażą się liczbami $ 1 $, $ q $, $ q^2 $. Aby istniał trójkąt, którego boki są odpowiednio równe tym odcinkom, potrzeba i wystarcza żeby suma każdych dwóch spośród tych odcinków była większa od trzeciego odcinka, tj. żeby spełnione były warunki: 1) $ 1 + q^2 > q $, 2) $ q + q^2 > 1 $, 3) $ 1 + q > q^2 $, którym możemy nadać postać:
Warunek 1) jest spełniony dla każdej wartości $ q $, gdyż wyróżnik trójmianu kwadratowego $ q^2 - q + 1 $ jest ujemny, $ \Delta = - 3 $, więc trójmian ma stale wartość dodatnią. Warunek 2) jest spełniony dla wartości $ q $ leżących po za pierwiastkami trójmianu $ q^2 + q - 1 $, tzn. gdy
Warunek 3) jest spełniony dla wartości $ q $ leżących między pierwiastkami trójmianu $ q^2 - q - 1 $, tzn. gdy
Ponieważ $ \frac{-1 - \sqrt{5}}{2} < \frac{1 - \sqrt{5}}{2} < \frac{-1 + \sqrt{5}}{2} < \frac{1 + \sqrt{5}}{2} $, więc ostatecznie otrzymujemy dla $ q $ warunek następujący:
Uwaga. Liczby $ \frac{- 1 + \sqrt{5}}{2} $ i $ \frac{1 + \sqrt{5}}{2} $ są odwrotnościami, gdyż ich iloczyn $ \frac{- 1 + \sqrt{5}}{2} \cdot \frac{1 + \sqrt{5}}{2} = 1 $. Stąd wynika, że jeśli pewna liczba $ q $ spełnia warunek (6), to jej odwrotność $ \frac{1}{q} $ też spełnia warunek (6) (rys. 3). Fakt ten był do przewidzenia; jeśli bowiem boki trójkąta wzięte w pewnej kolejności tworzą postęp geometryczny o ilorazie $ q $, to w odwrotnej kolejności dadzą one postęp geometryczny o ilorazie $ \frac{1}{q} $. Długość otrzymanego przedziału dla $ q $ wynosi $ \frac{1 + \sqrt{5}}{2} - \frac{- 1 + \sqrt{5}}{2} =1 $. Łatwo sprawdzić, ze liczby $ \frac{1 + \sqrt{5}}{2} $ i $ \frac{-1 + \sqrt{5}}{2} $ są jedyną parą liczb, które są odwrotnościami i których różnica równa się $ 1 $.
| \frac{-1+\sqrt{5}}{2}<\frac{1+\sqrt{5}}{2} | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 912 |
XL OM - III - Task 5
On a sphere of radius $ r $, there are three circles of radius $ a $, each tangent to the other two and contained within the same hemisphere. Find the radius of the circle lying on the same sphere and tangent to each of these three circles.
Note: Similar to on a plane, we say that two circles lying on a sphere are tangent if they have exactly one point in common. | Let's denote three given circles (of radius $a$) by $k_1$, $k_2$, $k_3$. Suppose $k_0$ is a circle of radius $x$, located on the same sphere and tangent to each of the circles $k_1$, $k_2$, $k_3$. None of the circles $k_1$, $k_2$, $k_3$ are great circles of the given sphere (two different great circles always have two points in common).
Let $O$ be the center of the sphere under consideration. The sum of the rays with a common origin $O$, intersecting the circle $k_i$, is the lateral surface of an unbounded circular cone, which we will denote by $C_i$: ($i = 0,1,2,3$).
The circle $k_0$ can be a great circle, the boundary of a hemisphere containing the circles $k_1$, $k_2$, $k_3$. In this case, the sum of the rays with origin $O$, determined by the points of the circle $k_0$, is a plane. Let's agree that by $C_0$ we mean in this case one of the two half-spaces determined by this plane - the one that contains the circles $k_1$, $k_2$, $k_3$. When we say "cone," we will continue to allow this case as well, treating the half-space as a limiting ("degenerate") form of a cone.
Take two different numbers $i, j \in \{0,1,2,3\}$. The circles $k_i$ and $k_j$ have exactly one common point, so the cones $C_i$ and $C_j$ have exactly one common generating line. We have two possibilities: either one of these cones is contained in the other, or the cones do not have any other common points besides the common generating line. In the first case, we will speak of internal tangency, in the second - of external tangency.
Notice that for each pair of numbers $i, j \in \{1,2,3\}$, the second case holds (given the assumed equality of the radii of the circles $k_1$, $k_2$, $k_3$, the inclusion $C_i \subset C_j$ would have to be an equality; the circles $k_1$ and $k_j$ would then be identical, contrary to the fact that they have only one point in common). Thus, the cones $C_1$, $C_2$, $C_3$ are pairwise externally tangent. The generating lines along which these cones are tangent are three different rays.
Geometric imagination reveals two configurations:
I. The cones $C_1$, $C_2$, $C_3$ are internally tangent to $C_0$.
II. All four cones are pairwise externally tangent.
(In case I, the circle $k_0$ "encompasses" the three given circles $k_1$, $k_2$, $k_3$, while in case II, it is "inscribed" between them.)
We will show that one of these situations must indeed occur.
Suppose this is not the case. Then at least one of the cones $C_1$, $C_2$, $C_3$ (for example, $C_1$) is internally tangent to $C_0$, and at least one of these cones (for example, $C_2$) is externally tangent to $C_0$. The third cone, $C_3$, is tangent to both $C_1$ and $C_2$ (along different generating lines). Some of its points must therefore be inside the cone $C_0$, and some - outside. This, however, cannot be reconciled with the tangency of $C_3$ to $C_0$. The obtained contradiction shows that there are indeed no other possibilities than I and II.
We proceed to the calculations.
Let the vertex angle of each of the cones $C_1$, $C_2$, $C_3$ be denoted by $2\alpha$, and the vertex angle of the cone $C_0$ by $2\varphi$:
Let $A_i$ be the center of the circle $k_i \in (i = 0,1, 2, 3)$. The triangle $A_1A_2A_3$ is equilateral. The length of its side will be determined by considering the section of the given solids and figures by the plane $OA_1A_2$ (Figure 7):
om40_3r_img_7.jpg
(on the figures, $\pi_i$ denotes the trace of the plane of the circle $k_i$).
Let $S$ be the center of the triangle $A_1A_2A_3$ (which is also the center of the inscribed circle and the circumscribed circle of this triangle). Its distance from the vertex equals $\frac{2}{3}$ of the height:
The segment $OA_1$ is the height of an isosceles triangle with two sides of length $r$ and an angle of $2\alpha$ between them. Therefore,
Now consider the section by the plane $OA_0A_1$. Let $B$ be the point of tangency of the circles $k_0$ and $k_1$.
om40_3r_img_8.jpg
om40_3r_img_9.jpg
We have the equalities: $|\measuredangle SOB| = \varphi$; $|A_1B| = a$, $|A_0B| = x$. The vertex angle of $\angle SOA_1$ is $\varphi - \alpha$ in case I and $\varphi + \alpha$ in case II (Figures 8 and 9 illustrate these two situations). Therefore, according to (1),
(sign minus in case I, plus in case II). At the same time, according to the relationships (2) and (3), we have the equality
By equating the right sides of (4) and (5), we get the equation
which, after squaring both sides and further transformations, takes the form
The roots of the quadratic equation (7) are the numbers
the expressions under the square root are non-negative, which follows from the relationship (5): $2a \leq \sqrt{3} r$. The number $x_1$ satisfies the radical equation (6) with the minus sign on the right side, and $x_2$ satisfies it with the plus sign.
Hence the answer:
For any triplet of pairwise tangent circles of equal radius $a$ (on a hemisphere of radius $r$), there exist two circles tangent to each of these three circles, arranged as described above (configurations I and II). Their radii are given by the formulas (8); the value $x_1$ corresponds to configuration I, and the value $x_2$ corresponds to configuration II. | Geometry | math-word-problem | Yes | Yes | olympiads | false | 915 | |
XXII OM - II - Problem 4
In the plane, a finite set of points $ Z $ is given with the property that no two distances between points in $ Z $ are equal. Points $ A, B $ belonging to $ Z $ are connected if and only if $ A $ is the nearest point to $ B $ or $ B $ is the nearest point to $ A $. Prove that no point in the set $ Z $ will be connected to more than five other points. | Suppose that point $ A \in Z $ is connected to some points $ B $ and $ C $ of set $ Z $ and let, for example, $ AB < AC $. Then point $ C $ is not the closest to $ A $, and from the conditions of the problem it follows that point $ A $ is the closest to $ C $. Therefore, $ AC < BC $. Hence, in triangle $ ABC $, side $ \overline{BC} $ is the longest, and thus angle $ \measuredangle BAC $ is the largest. It follows that $ \measuredangle BAC > \frac{\pi}{3} $.
We have thus proved that if point $ A \in Z $ is connected to some points $ B $ and $ C $ of set $ Z $, then $ \measuredangle BAC > \frac{\pi}{3} $.
If, therefore, point $ A \in Z $ is connected (Fig. 13) to points $ B_1, B_2, \ldots, B_n $ of set $ Z $, then
Adding these inequalities side by side, we obtain that $ 2\pi > n \frac{\pi}{3} $, i.e., $ n < 6 $. Therefore, the number $ n $ of points of set $ Z $ to which point $ A $ is connected is not greater than $ 5 $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 917 |
LII OM - III - Problem 6
Given positive integers $ n_1 < n_2 < \ldots < n_{2000} < 10^{100} $. Prove that from the set $ \{n_1, n_2, \ldots, n_{2000}\} $, one can select non-empty, disjoint subsets $ A $ and $ B $ having the same number of elements, the same sum of elements, and the same sum of squares of elements. | For a set $ X \subseteq \{n_1,n_2,\ldots,n_{2000}\} $, let $ s_0(X) $, $ s_1(X) $, and $ s_2(X) $ denote the number of elements, the sum of elements, and the sum of squares of elements of the set $ X $, respectively.
It suffices to prove that there exist two different subsets $ C $ and $ D $ of the set $ \{n_1,n_2,\ldots,n_{2000}\} $ such that $ s_i(C) = s_i(D) $ for $ i = 0,1,2 $. Then the sets $ A = C \setminus D $ and $ B = D\setminus C $ are non-empty and satisfy the conditions of the problem.
For any subset $ X $ of the set $ \{n_1 , n_2, \ldots ,n_{2000}\} $ we have
Denoting $ S(X) = s_0(X) + 10^4s_1(X) + 10^{108}s_2(X) $, we obtain
Hence, there exist two different subsets $ C $, $ D $ of the set $ \{n_1 , n_2, \ldots ,n_{2000}\} $ such that $ S(C) = S(D) $. From inequality (1), it follows that $ s_i(C) = s_i(D) $ for $ i = 0, 1, 2 $, which completes the solution of the problem. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 920 |
XLVIII OM - II - Problem 6
In a cube with edge length $1$, there are eight points. Prove that some two of them are the endpoints of a segment of length not greater than $1$. | Let the vertices of a given cube $\mathcal{C}$ be denoted by $A_1, \ldots, A_8$. Let $\mathcal{C}_i$ be the cube with edge length $\frac{1}{2}$, having one vertex at point $A_i$ and three faces contained in the faces of cube $\mathcal{C}$. Let $P_1, \ldots, P_8$ be eight given points.
Each cube $\mathcal{C}_i$ has a diameter of $\frac{1}{2}\sqrt{3} < 1$. Therefore, if two different points $P_k$ and $P_l$ lie in any of the cubes $\mathcal{C}_i$, then $P_kP_l < 1$; the required condition is satisfied.
Assume, then, that each of these cubes contains exactly one point $P_i$. Fix the numbering so that $P_i \in \mathcal{C}_i$ for $i = 1, \ldots, 8$. Each cube $\mathcal{C}_i$ has exactly three faces in common with the surface of cube $\mathcal{C}$; the point $P_i$ is at a distance of no more than $\frac{1}{2}$ from each of these three faces. Denote these three distances by $a_i$, $b_i$, $c_i$. Let $d$ be the largest of the 24 numbers: $a_1, b_1, c_1, \ldots, a_8, b_8, c_8$. Without loss of generality, we can assume that $d = P_1Q$, where $Q$ is the orthogonal projection of point $P_1$ onto a certain face of cube $\mathcal{C}$, and that $A_1A_2$ is an edge of cube $\mathcal{C}$ parallel to the line $P_1Q$.
Let $R$ be the orthogonal projection of point $P$ onto the line $A_1A_2$. Consider the rectangular prism whose one edge is the segment $A_2R$, and one of the faces perpendicular to $A_2R$ is a square with vertex $A_2$ and side length $d$, contained in a face of cube $\mathcal{C}$. From the definition of the number $d$, it follows that this rectangular prism contains points $P_1$ and $P_2$. Its diameter is equal to
since $0 \leq d \leq \frac{1}{2} < \frac{2}{3}$. Therefore, $P_1P_2 \leq 1$, which completes the proof. | proof | Geometry | proof | Yes | Yes | olympiads | false | 921 |
L OM - I - Problem 1
Prove that among numbers of the form $ 50^n + (50n + 1)^{50} $, where $ n $ is a natural number, there are infinitely many composite numbers. | I way:
If $ n $ is an odd number, then the number $ 50^n $ when divided by $ 3 $ gives a remainder of $ 2 $. If, moreover, $ n $ is divisible by $ 3 $, then $ (50n+1)^{50} $ when divided by $ 3 $ gives a remainder of $ 1 $. Hence, the number $ 50^n + (50n+1)^{50} $ is divisible by $ 3 $ for numbers $ n $ of the form $ 6k + 3 $.
II way:
For $ n $ divisible by $ 5 $, the number $ 50^n +(50n+1)^{50} $ is the sum of fifth powers. Therefore, this number, by virtue of the identity
is composite. | proof | Number Theory | proof | Yes | Yes | olympiads | false | 922 |
XXXIII OM - II - Problem 2
The plane is covered with circles in such a way that the center of each of these circles does not belong to any other circle. Prove that each point of the plane belongs to at most five circles. | Suppose that point $ A $ belongs to six circles from the considered family and that $ O_1 $, $ O_2 $, $ O_3 $, $ O_4 $, $ O_5 $, $ O_6 $ are the centers of these circles. It follows that one of the angles $ O_iAO_j $ $ (i,j = 1, 2, \ldots, 6) $ has a measure not greater than $ 60^\circ $ (six of the angles $ O_iAO_j $ sum up to a full angle), hence in the triangle $ O_iAO_j $ the side $ O_iO_j $ opposite this angle is not the longest side, since $ O_iO_j \leq \max (AO_i,AO_j) $. Assume that $ O_iO_j \leq AO_i $. It follows that: $ O_j $ belongs to the circle with center $ O_i $. This contradicts the assumption that the center of each of the given circles is not a point of any other circle. Therefore, there are no points in the plane belonging to more than five circles of the given family. | proof | Geometry | proof | Yes | Yes | olympiads | false | 923 |
LVII OM - II - Problem 5
Point $ C $ is the midpoint of segment $ AB $. Circle $ o_1 $ passing through points $ A $ and $ C $ intersects circle $ o_2 $ passing
through points $ B $ and $ C $ at different points $ C $ and $ D $. Point $ P $ is the midpoint of the arc $ AD $ of circle $ o_1 $, which does not
contain point $ C $. Point $ Q $ is the midpoint of the arc $ BD $ of circle $ o_2 $, which does not contain point $ C $. Prove that the lines $ PQ $ and $ CD $
are perpendicular. | If $ AC = CD $, then also $ BC = CD $. Then segments $ PC $ and $ QC $ are diameters of circles $ o_1 $ and $ o_2 $, respectively. Therefore, $ \measuredangle CDP = \measuredangle CDQ = 90^{\circ} $, which implies that point $ D $ lies on segment $ PQ $, and lines $ PQ $ and $ CD $ are perpendicular.
Let us assume in the further part of the solution that $ AC \neq CD $.
Let $ E $ be a point lying on ray $ CD $ such that $ CE = AC $ (Fig. 3). Then also $ CE = BC $.
From the relationships $ CE = AC $ and $ \measuredangle ACP = \measuredangle ECP $, it follows that triangles $ ACP $ and $ ECP $ are congruent
(side-angle-side criterion). Therefore, $ EP = AP = DP $.
Similarly, we prove that $ EQ = DQ $.
om57_2r_img_3.jpg
From the obtained equalities, it follows that points $ D $ and $ E $ are symmetric with respect to line $ PQ $. Line $ DE $ is therefore perpendicular
to line $ PQ $, which completes the solution of the problem. | proof | Geometry | proof | Yes | Yes | olympiads | false | 924 |
XVIII OM - II - Task 2
In a room, there are 100 people, each of whom knows at least 66 of the remaining 99 people. Prove that it is possible that in every quartet of these people, there are two who do not know each other. We assume that if person $ A $ knows person $ B $, then person $ B $ also knows person $ A $. | Let's denote the people in the room by the letters $A_1, A_2, \ldots, A_{100}$. Let $M$ be the set of people $\{A_1, A_2, \ldots, A_{33}\}$, $N$ be the set of people $\{A_{34}, A_{35}, \ldots, A_{66}\}$, and $P$ be the set of people $\{A_{67}, A_{68}, \ldots, A_{100}\}$. The case described in the problem occurs, for example, when each person in set $M$ knows only and all the people in sets $N$ and $P$ (a total of 67 people), and similarly, each person in set $N$ knows all the people in sets $P$ and $M$ and only those people (67 people), and each person in set $P$ knows all the people in sets $M$ and $N$ and only those people (66 people). If ($A_i, A_j, A_k, A_l$) is any quartet of these people, then two of them are in the same set $M$, $N$, or $P$, and therefore, these two people do not know each other.
Note. The theorem above can be easily generalized. Suppose there are $N$ people in the room, each of whom knows at least $\left[ \frac{2n}{3} \right]$ of the other people. It can then happen that in every quartet of these people, there are two who do not know each other. To prove this, divide the set $M$ of all people into 3 sets $M_1$, $M_2$, and $M_3$ containing $m_1 = \left[ \frac{n+1}{3} \right]$, $m_2 = \left[ \frac{n+1}{3} \right]$, and $m_3 = n - 2 \left[ \frac{n+1}{3} \right]$ people, respectively. If each person in set $M_i$ knows the people in sets $M_j$ and $M_k$ ($i, j, k = 1, 2, 3; i \ne j \ne k \ne i$) and only those people, then in every quartet of people in set $M$, there are two people who belong to the same set $M_i$, so they do not know each other. Each person in set $M$ then knows at least $\left[ \frac{2n}{3} \right]$ people. This is easiest to verify separately for each of the three possible cases: $n = 3k$, $n = 3k+1$, $n = 3k + 2$.
Namely
We see from this that the given inequalities are satisfied for any natural number $n$. | proof | Combinatorics | proof | Yes | Yes | olympiads | false | 925 |
LV OM - I - Task 7
Find all solutions to the equation $ a^2+b^2=c^2 $ in positive integers such that the numbers $ a $ and $ c $ are prime, and the number $ b $ is the product of at most four prime numbers. | There are three solutions $(a,b,c): (3,4,5), (5,12,13), (11,60,61)$. Let $a$, $b$, $c$ be numbers satisfying the conditions of the problem. Then
from the assumption that the number $a$ is prime, we obtain $c = b + 1$. Therefore, $a^2 = 2b + 1$, which implies that the number $a$ is odd.
Let $a = 2n + 1$. Then
if the number $n$ gives a remainder of 1 when divided by 3, then the number $a$ is divisible by 3. Therefore, it is composite, except for the case $n = 1$, which gives the solution $(a,b,c) = (3,4,5)$. If, on the other hand, the number $n$ gives a remainder of 2 when divided by 5, then the number $a$ is divisible by 5. Therefore, it is composite, except for the case $n = 2$. This case gives the second solution $(a,b,c) = (5,12,13)$.
If, however, the number $n$ gives a remainder of 1 or 3 when divided by 5, then the number $c$ is divisible by 5. Therefore, it is composite except for the case $n = 1$, which we have already considered.
There remains the case where the number $n$ gives a remainder of 0 or 2 when divided by 3, and a remainder of 0 or 4 when divided by 5. In this case, the number $b$ is divisible by $2 \cdot 6 \cdot 5 = 60$. On the other hand, the number $b$ is the product of at most four prime numbers, which implies that $b = 60$. This case leads to the third solution: $(a,b,c) = (11,60,61)$. | (3,4,5),(5,12,13),(11,60,61) | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 926 |
XLV OM - III - Task 2
In the plane, there are two parallel lines $ k $ and $ l $ and a circle disjoint from line $ k $. From a point $ A $ lying on line $ k $, we draw two tangents to this circle intersecting line $ l $ at points $ B $, $ C $. Let $ m $ be the line passing through point $ A $ and the midpoint of segment $ BC $. Prove that all such lines $ m $ (corresponding to different choices of point $ A $ on line $ k $) have a common point. | Note 1. The conditions of the problem do not change if the given line $ l $ is replaced by another line $ l' $ parallel to $ k $ and different from $ k $. Indeed: if two tangents to a given circle emanating from point $ A $ intersect the line $ l $ at points $ B $ and $ C $, then the midpoints of segments $ BC $ and $ B $ are collinear with point $ A $, and thus the line $ m $ defined in the problem coincides with the line $ m' $ obtained in the same way after replacing the line $ l $ with the line $ l' $. We can therefore assume that $ l $ is a line parallel to $ k $, chosen in a convenient position for the considerations being made.
Let $ I $ and $ r $ be the center and radius of the given circle $ \Omega $, and let $ N $ and $ S $ be the points of this circle that are, respectively, closest and farthest from the line $ k $. In light of the observation just made, we can assume that the line $ l $ is tangent to the circle $ \Omega $ at point $ S $. Let the distance between the lines $ k $ and $ l $ be denoted by $ h $.
Fix a point $ A $ on the line $ k $. Let $ B $ and $ C $ be the points on the line $ l $ determined according to the problem statement. The circle $ \Omega $ is inscribed in the triangle $ ABC $. Denote the midpoint of side $ BC $ by $ M $, and the foot of the perpendicular from vertex $ A $ by $ D $ (so $ |AD| = h $). Without loss of generality, we can assume that $ |AB| \leq |AC| $. Then the points $ B $, $ S $, $ M $, $ C $ lie on the line $ l $ in that order (in a special case, points $ S $ and $ M $ may coincide). Figure 13 shows the situation when angle $ ABC $ is acute; but the considerations (and calculations) that follow also cover the case when this angle is right or obtuse.
om45_3r_img_13.jpg
Since $ S $ is the point of tangency of the inscribed circle in triangle $ ABC $ with side $ BC $, the length of segment $ BS $ is given by the known formula
\[
BS = \frac{a + c - b}{2}
\]
where, as usual, $ a = |BC| $, $ b = |CA| $, $ c = |AB| $. Therefore,
\[
BS = \frac{a + c - b}{2}
\]
Further, denoting by $ d $ the distance between points $ D $ and $ M $, we have the equalities
\[
d = \frac{1}{2}a - \frac{a + c - b}{2} = \frac{b - c}{2}
\]
(The sign of the difference $ \frac{1}{2}a - d $ depends on whether angle $ ABC $ is acute or not). From this (and the relationships in the right triangles $ ABD $ and $ ACD $), we get the relationship
\[
h^2 = AD^2 = AM^2 - DM^2 = \left(\frac{a}{2}\right)^2 - \left(\frac{b - c}{2}\right)^2 = \frac{a^2 - (b - c)^2}{4}
\]
Thus,
\[
h^2 = \frac{a^2 - (b - c)^2}{4}
\]
A useful equality is $ ah = (a + b + c)r $, both sides of which express twice the area of triangle $ ABC $. It follows that $ r(b + c) = a(h - r) $, or
\[
r(b + c) = a(h - r)
\]
Let $ X $ be the point of intersection of the line $ m $ (i.e., $ AM $) with the line $ NS $. From the similarity of triangles $ MDA $ and $ MSX $, we have the proportion $ |SX|:|SM| = |AD| : |MD| $. Considering the relationships (2), (3), (4), we obtain the equality
\[
|SX| = \frac{h \cdot |SM|}{|MD|} = \frac{h \cdot \frac{a}{2}}{\frac{b - c}{2}} = \frac{ha}{b - c}
\]
The last quantity depends only on $ h $ and $ r $; it is therefore determined solely by the line $ k $ and the circle $ \Omega $, and does not depend on the choice of point $ A $. This means that the point $ X $, located on the segment $ SN $ at a distance $ \frac{hr}{h - r} $ from point $ S $, is the common point of all the lines $ m $. | proof | Geometry | proof | Yes | Yes | olympiads | false | 927 |
XIV OM - II - Task 4
In triangle $ABC$, the angle bisectors of the internal and external angles at vertices $A$ and $B$ have been drawn. Prove that the perpendicular projections of point $C$ onto these bisectors lie on the same line. | Let $M$ and $N$ denote the projections of point $C$ onto the angle bisector $AM$ of angle $A$ of triangle $ABC$ and onto the bisector of the adjacent angle (Fig. 22).
Since the bisectors $AM$ and $AN$ are perpendicular, the quadrilateral $AMCN$ is a rectangle. The point $S$, where the diagonals $MN$ and $AC$ intersect, is the midpoint of $AC$. Notice that the line $NM$ is symmetric to the line $AC$ with respect to the line $ST$ parallel to $AM$, and the line $AC$ is symmetric to the line $AB$ with respect to the line $AM$. Successive applications of mirror reflections (symmetries) with respect to the parallel lines $ST$ and $AM$ transform the line $NM$ into the line $AB$.
Since the composition of two mirror reflections of a figure with respect to two parallel axes is a translation of that figure, the line $NM$ is parallel to the line $AB$. The line $NM$ is thus the line passing through the midpoints of sides $AC$ and $BC$. The same applies to the line $PQ$ drawn through the projections $P$ and $Q$ of point $C$ onto the angle bisector of angle $B$ of triangle $ABC$ and onto the bisector of the adjacent angle. Therefore, the lines $MN$ and $PQ$ coincide, q.e.d.
Note. In the above solution, we have relied on an important theorem in plane geometry, that the transformation of a figure consisting of the successive application of two mirror reflections (axial symmetries) with respect to two parallel lines is a translation of that figure. The proof of this theorem is very simple.
Let $a$ and $b$ be two parallel lines, and $\overline{W}$ the vector of this translation parallel to $a$, which transforms $a$ into $b$. Let $P$ be any point in the plane, $Q$ the point symmetric to $P$ with respect to line $a$, $R$ the point symmetric to $Q$ with respect to line $b$, and finally let $A$ and $B$ be the midpoints of segments $PQ$ and $QR$ (Fig. 23).
Then, the point $R$ is obtained by translating the point $P$ in a direction perpendicular to the lines $a$ and $b$ by a distance equal to twice the distance between the lines $a$ and $b$.
Such a translation will be experienced by any figure when it is first transformed by symmetry with respect to line $a$ and then with respect to line $b$. Conversely, any translation can be replaced by the composition of two axial symmetries with respect to axes perpendicular to the direction of the translation, whose distance is half the length of the translation. One of these axes can be drawn through any point in the plane.
Equally important is the theorem:
The transformation of the plane consisting of the composition of two axial symmetries with respect to axes $a$ and $b$ intersecting at point $O$ is a rotation around point $O$ by an angle twice the angle between $a$ and $b$.
The proof, illustrated by Fig. 24, is analogous to the previous one: ,
Conversely, any rotation in the plane can be replaced by the composition of two mirror reflections with respect to two axes passing through the center of rotation, whose angle is half the angle of rotation. One of these axes can be drawn through any point in the plane. | proof | Geometry | proof | Yes | Yes | olympiads | false | 928 |
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